IN   MEMORIAM 
FLORIAN  CAJOR1 


bi 


RECOMMENDATIONS. 


From  Rev.  C  H.  Alden,  Principal  of  the  Philadelphia  High  School  for 

Young  Ladies. 
Mr.  Green  : 

Dear  sir, — I  am  greatly  pleased  to  find,  that,  in  your  "  Gradations  in 
Algebra,"  recently  published,  you  have  rendered  the  elements  of  that  de- 
partment of  the  Mathematics  so  attractive  to  the  young  student.  It  has 
long  surprised  me  that  this  interesting  method  of  analysis  has  been  so 
entirely  excluded  from  our  common  schools.  With  your  valuable  aid, 
however,  this  neglect  can  no  longer  find  a  suitable  apology.  I  anticipate 
the  introduction  of  your  excellent  work  into  both  our  private  primary  and 
public  schools. 

Very  respectfully, 

C.  H.  ALDEN. 


Richard  W.  Green: 

Sir, — I  have  examined  your  "  Gradations  in  Algebra"  with  much  care, 
and  have  no  hesitation  in  saying  that  in  my  opinion  you  have  fully  attained 
your  object  in  forming  "  an  easy  introduction  to  the  first  principles  of  alge- 
braical reasoning,"  and  of  furnishing  "in  the  same  course  a  popular  expo- 
sition of  the  most  important  elements  of  arithmetic."  To  write  a  work  on 
abstruse  science,  adapted  to  the  comprehension  of  youth,  is  a  work  of 
extreme  difficulty.  I  feel  that  I  offer  high  but  deserved  and  just  praise 
when  I  say  that  in  the  book  before  me  you  have  fully  succeeded. 

Yours,  «fec. 

JAS.  RHODES, 
Principal  of  N.  W.  Grammar  School,  Philad. 


Dear  sir, — Your  "  Gradations  in  Algebra"  appears  to  me  to  supply 
what  has  long  been  wanted  by  those  who  are  commencing  that  study,  viz. 
an  initiatory  text-book  at  once  small,  clear,  well  arranged,  and  compre- 
hensive. It  meddles  with  nothing  beyond  the  capacity  of  a  school-boy. 
It  is  well  supplied  with  such  examples  as  show  the  learner  the  use  of  what 
he  is  studying.  I  would  rather  put  your  book  in  the  hands  of  a  beginner 
than  any  other  work  on  the  same  subject  now  in  common  use. 

Yours  very  respectfully, 

JNO.  W.  FAIRES. 
A  i 


D  RECOMMENDATIONS. 

I  examined  "  Green's  Gradations  in  Algebra,"  and  was  so  well  pleased 
with  it,  that  I  introduced  it  into  our  school.  I  have  made  use  of  it  one 
term,  and  can  bear  testimony  that  it  has  equaled  my  expectations.  I  con- 
sider it  as  occupying  an  important  link  between  arithmetic  and  literal  alge- 
bia.     In  my  judgment,  we  have  long  needed  such  a  work. 

JOHN  D.  POST, 
Teacher  of  Mathematics  in  Hartford  Gr  xmmar  School. 


From  Porter  H.  Snow,  A.  M.,  formerly  Principal  of  the  Hartford  Centre 
School,  and  now  Principal  of  Brainard  Academy,  Haddam,  Conn. 

I  have  examined  critically  your  "Gradations  in  Algebra,"  and  find  it 
admirably  adapted  to  fulfil  its  design.  There  is  nothing  so  good  on  the 
subject  in  the  English  language  for  a  student  to  read  before  entering  col- 
lege. All  the  mystery  that  clothes  mathematics  during  the  early  part  of  a 
collegiate  course  of  study  will  be  solved  and  made  plain,  if  the  student 
will  spend  a  few  weeks  on  this  introduction  to  algebra.  It  is  also  a  fine 
work  for  classes  in  schools.  And  I  would  recommend  its  introduction  into 
primary  and  higher  schools  in  preference  to  any  treatise  on  algebra  I  have 
ever  seen.  It  is  also  the  only  thing  of  the  kind  I  have  met  with,  suitable 
for  a  student  to  peruse  without  an  instructor. 


York,  Pa.,  Nov.  30th,  1841. 
After  a  careful  examination  of  Mr.  Green's  "Algebra,"  we  have  no 
hesitation  in  expressing  our  decided  approbation  of  the  work.  The  clear- 
ness of  the  explanations,  the  judicious  and  systematic  arrangement  of  the 
parts,  and  the  gradual  manner  in  which  the  student  is  led  on  from  the  first 
principles  to  the  more  difficult  parts  of  the  science,  render  it,  in  our  view, 
preferable  to  any  we  have  seen  for  primary  instruction. 

DANIEL  KIRKWOOD, 
Teacher  of  Mathematics,  York  Co.  Academy. 
D.  M.  ETTINGER, 
Principal  of  the  High  School,  York,  Pa. 


The  great  excellence  of  this  work  is,  that  it  brings  the  pupil  on  gradually. 
Only  one  difficulty  is  presented  at  a  time  ;  and  upon  this,  explanations  and 
examples  are  multiplied  i ill  the  pupil  becomes  so  familiar  with  the  subject 
that  he  almost  wonders  where  the  difficulty  is  at  which  he  first  stumbled. 
The  examples  are  numerous  ;  the  few  rules  (and  they  are  enough)  are  ad- 
mirably expressed,  in  plain,  concise  language.  On  the  whole,  we  may  say, 
that  U  a  vouth  is  to  begin  the  study  of  algebra,  this  should  be  his  first  book 

Extract  from  the  North  American. 


GRADATIONS 

IN 

ALGEBRA, 

IN    WHICH 

THE   FIRST   PRINCIPLES   OF   ANALYSIS 

ARE 

INDUCTIVELY  EXPLAINED. 

ILLUSTRATED   BY 
COPIOUS    EXERCISES, 

AND  MADE  SUITABLE  FOP  FRIM^RY  SCHOOLS. 

BY 

RICHARD    W.    GREEN,    A.M., 

II 

AUTHOR  OF  ARITHMETICAL  GUIDE,  LITTLE  RECKONER,  ETC. 


PHILADELPHIA: 

PUBLISHED  BY  E.  H.  BUTLER  &  CO. 

1850. 


QhiS3 

£7 


(  Chamber  of  the  Controllers  of  Public  Schools, 
(  First  School  District  of  Pennsylvania. 

Philadelphia,  November  15,  1849. 
At  a  meeting  of  the  Controllers  of  Public  Schools,  First  District  of  Penn 
sylvania,  held  at  the  Controllers'  Chamber,  on  Tuesday,  November  13, 
1849,  the  following  Resolution  was  adopted  : — 

Resolved,  That  Green's  Algebra  be  introduced,  as  a  Class-Book,  into 
the  Grammar  Schools  of  the  District.  From  the  minutes, 

R.  J.  HEMPHILL,  Secretary. 

Secretary's  Office,  Harrisburg,  Feb.  4,  1843. 
Mr.  Richard  W.  Green  : 
Sir, — I  have  examined  the  work  prepared  by  you  for  the  use  of  primary 
schools,  entitled  "Gradations  in  Algebra."  It  meets  with  my  approba- 
tion, and  I  am  much  pleased  with  its  design  and  arrangement.  I  consider 
it  admirably  calculated  to  aid  the  pupil  at  the  commencement  of  the 
science,  also  to  give  him  a  general  knowledge  of  it. 

It  is,  in  my  opinion,  well  adapted  to  the  use  of  common  schools,  where 
there  are  students  who  have  not  the  time  or  means  of  consulting  more  ex- 
tended treatises  upon  the  subject.  I  hope  it  may  be  introduced  generally 
into  our  schools  throughout  the  state. 

I  am  yours  respectfully, 

A.  V.  Parsons, 
\  .'    tSfyjptrintqndent  Common  Schools. 


Entered  according  la  the  Act  of  Congress,  in  the  year  1^39,  by 

Richard  W.  Green, 

in  the  Clerk's  Office  of  the  District  Court  of  ihe  Eastern  District  of 

Pennsylvania. 

CAJORI 

$3"  For  the  convenience  of  those  teachers  who  may  wish  to  adopt  the 
author's  plan  of  using  this  book,  a  key  has  been  prepared  containing  the 
solutions  of  all  the  examples  and  problems.  It  will  also  save  the  teacher 
much  time  when  he  wishes  to  find  for  his  pupil  the  errors  in  his  v»  ork. 


CONTENTS 


Preface page  7 

NUMERAL    ALGEBRA. 

Preliminary  Remarks 11 

Addition  and  Subtraction  of  Simple  Quantities 18 

General  Rule  for  uniting  Terms 19 

Multiplication  and  Division  of  Simple  Quantities 23 

Simple   Equations 16 

I.  Equations  Solved  by  uniting  Terms 24 

II.  Addition  of  Compound  Quantities 22 

Transposition  by  Subtraction 30 

Equations  Solved  by  Transposition 31 

III.  Transposition  by  Addition 34 

Equations 34 

IV.  Transposition  of  the  Unknown  Quantity 37 

Equations 38 

V.  Multiplication  of  Compound  Quantities  by  Simple  Quantities. ...  42 
Equations 44 

VI.  Fractions 48 

Equations 52 

VII.  Fractions  of  Compound  Quantities 57 

Equatl  >ns 58 

VIII.  Division  of  Fractions  and  Fractions  of  Fractions 61 

Equations 64 

IX.  Subtraction  of  Compound  Quantities 67 

Equations 73 

X.  Uniting  Fractions  of  different  denominators 69 

Ratio  and  Proportion 75 

Equations 78 


9185-13 


VI  CONTENTS. 

XI.  Equations  with  two  Unknown  Quantities Page  82 

First  method  of  Extermination ." 83 

Equations 88 

XII.  Second  method  of  Extermination 93 

Equations    94 

XIII.  Third  method  of  Extermination 97 

Equations 98 

XIV.  Equations  with  several  Unknown  Quantities 101 

LITERAL    ALGEBRA. 

General  Principles 103 

Addition  and  Subtraction  of  Algebraical  quantities 112 

Multiplication  of  Algebraical  Quantities 117 

General  Properties  of  N  umbers 125 

Division  of  Algebraical  Quantities 128 

Division  by  Compound  Divisors 136 

Reduction  of  Fractions  to  lower  Terms .• 143 

Multiplication  where  one  factor  is  a  fraction 146 

Reducing  Complex  Fractions  to  Simple  ones 148 

Division  of  Fractions 152 

Fractions  of  Fractions 153 

Uniting  Fractions  of  different  denominators 156 

Division  by  Fractions 159 

General  Theory  of  Equations  with  two  unknown  quantities 162 

Involution  and  powers 1 65 

Evolution 172 

Extractioti  of  the  Second  Root  of  Numbers 177 

XV.  Pure  Quadratic  Equations 181 

Equations 182 

XVI.  Affected  Quadratic  Equations 185 

Equations 1 90 


PREFACE 


The  object  of  the  author,  in  composing  this  treatise,  was 
to  form  an  easy  introduction  to  the  first  principles  of  alge- 
braical reasoning ;  and  also  to  embrace,  in  the  same  course, 
a  popular  exposition  of  the  most  important  elements  of 
arithmetic.  And  he  believes  that  he  has  been  enabled  to 
combine  the  rudiments  of  both,  in  such  a  manner  as  to  make 
the  operations  of  one  illustrate  the  principles  of  the  other. 

In  order  that  this  method  of  treating  the  subject  might 
preserve  its  chief  advantage,  especially  in  the  initiatory  course 
of  the  study ;  the  work  has  been  divided  into  two  parts— 
Numeral  Algebra  and  Literal  Algebra. 

In  Numeral  Algebra  I  have  treated  of  the  several  primary 
arithmetical  operations;  first  making  them  intelligible  to  very 
youngf  pupils,  and  then  exhibiting  them  under  the  algebraical 
notation.  By  this  means,  as  every  lesson  in  algebra  is  imme- 
diately preceded  by  corresponding  numerical  exercises,  the 
transition  from  one  to  the  other  has  been  made  so  trifling, 
that  the  pupil  will  feel  at  each  step  that  he  has  met  with 
nothing  more  than  what  he  has  already  made  himself  familiar 
with  in  a  different  dress. 

Besides,  as  algebraical  operations  require  the  exercise  of 
abstraction  in  a  greater  degree  than  the  pupil  is  supposed  to 
be  accustomed  to,  I  have  taken  care  that  the  exercise  on  each 
of  the  fundamental  rules,  shall  be  followed  by  a  selection  of 
problems  to  be  solved  by  equations. 

7 


8  PREFACE. 

As  mathematical  questions  of  this  kind  are  always  pleas- 
ing to  young  pupils,  this  arrangement  will  serve  to  impart 
an  interest  to  the  study  at  the  commencement,  and  also  to 
preserve  a  taste  for  it  through  the  whole  course. 

Indeed,  this  part  of  the  work  is  the  most  important  for 
those  pupils  who  do  not  intend  to  pursue  the  mathematical 
sciences.  For,  it  is  in  such  exercises,  that  the  mind  is 
trained  to  investigate  the  relations  of  one  thing  to  another,  and 
to  conduct  its  reasonings  in  a  clear  and  forcible  manner. 

Under  the  head  of  Literal  Algebra,  I  have  repeated,  in 
a  more  strictly  algebraical  form,  the  principles  which  have 
been  explained  in  the  preceding  part  of  the  work ;  and  have 
shown  some  of  their  uses  by  applying  them  in  the  deduc- 
tion and  demonstration  of  several  abstruse  operations  on 
numbers. 

But  the  great  peculiarity  of  the  book  is,  that  it  habituates 
the  speech  and  the  ear  to  mathematical  language.  In  any 
study,  it  is  necessary  for  beginners  to  receive  such  a  course 
of  training  as  will  imprint  upon  their  minds  each  new  idea, 
as  soon  as  it  is  apprehended.  Learners  in  the  mathematics, 
especially,  are  accustomed  to  forget  soon,  both  the  names 
and  the  use  of  the  signs;  and  also  the  arrangement  of  the 
several  steps  in  the  solution  of  their  problems.  On  this 
account  I  have  required  the  pupil  always  to  repeat  verbally 
the  operation  that  he  has  performed ;  taking  care  to  omit  no 
part  of  the  work  that  would  hinder  an  auditor  from  under- 
standing the  reason  for  the  several  steps,  and  consenting  to 
the  just  conclusions  of  the  answer  which  has  been  obtained. 
It  has  been  found  by  experience  that  this  simple  device 
enables  the  young  pupil  to  acquire  the  science  very  easily; 


PREFACE.  9 

and  while  it  impresses  his  lessons  indelibly  upon  his 
memory,  it  also  developes  his  genius,  rectifies  his  inventive 
faculties,  and  imparts,  as  it  were,  a  mathematical  form  to  his 
mind ;  so  much  so,  that  he  is  generally  capable  of  pursuing 
the  subject  afterwards  by  himself. 

In  order  to  accomplish  this  end  more  perfectly,  I  have 
swelled  the  number  of  examples  beyond  the  ordinary  limits. 
These  should  be  thoroughly  mastered  as  the  pupil  proceeds. 
There  must  be  no  smattering  in  the  beginning  of  a  science 
if  the  learner  is  to  continue  the  study.  The  author  has 
found  by  long  experience,  that  a  book  is  sooner  finished 
when  each  part  has  been  made  familiar  to  the  mind,  than 
when  it  has  been  superficially  attended  to. 

With  regard  to  the  arrangement  of  the  several  divisions, 
I  have  been  careful  to  introduce  first  those  principles  that 
will  be  the  most  easily  apprehended ;  and  afterwards  such 
others  as  would  most  naturally  arise  from  the  former  if  the 
study  were  entirely  new.  This  method  appears  to  be  the 
best  adapted  for  teaching  the  rudiments  of  a  science ; 
although  in  a  succeeding  text  book,  it  is  necessary  that  the 
arrangement  of  the  several  parts  should  be  more  systematic. 
On  this  account  the  advanced  scholar  must  not  be  surprised 
to  find  in  the  middle  of  the  book,  what  he  has  been  accus- 
tomed to  see  near  the  beginning  of  other  treatises.  How- 
ever, so  much  regard  to  a  regularity  of  arrangement  has 
been  attended  to,  that  the  pupil  will  be  assisted  by  the  asso- 
ciations of  method,  both  to  understand  and  to  remember. 

As  the  author  wishes  to  bring  the  study  of  Algebra  withir 
the  reach  of  common  schools,  he  has  endeavored  to  pre 
pare  this  work,  so  that  it  may  be  studied  by  pupils  who  are 


10  PREFACE. 

not   already  adepts   in  arithmetic.     And  it  is  believed   lhat 

such  learners  will  not  fail  of  obtaining,  by  a  perusal  of  it, 

a  full  understanding  of  vulgar  fractions,  roots  and  powers, 

proportion,  progression,  and  other  numerical  operations  that 

are  generally  embraced  in  arithmetical  treatises. 

R.  W.  G. 


ADVERTISEMENT. 

The  foregoing  is  the  Preface  of  the  author's  "  Inductive 
Algebra."  The  first  192  pages  of  that  book  have  been 
published  in  this  form,  in  order  to  afford  a  cheap  manual 
for  those  classes  that  do  not  wish  to  study  beyond  Quadratic 
Equations.  In  the  present  state  of  education,  so  much  of 
Algebra  should  be  studied  by  every  pupil  in  our  common 
schools 


PART   L 


NUMERAL    ALGEBRA. 

PRELIMINARY    REMARKS. 

§  1.  Algebra  is  a  method  of  arithmetical  computation, 
in  which  the  calculations  are  performed  by  means  of  letters 
and  signs. 

§  2.  When  the  answer  of  an  arithmetical  question  is  to 
be  found  by  algebra,  we  first  represent  that  answer  by  a 
letter;  because  we  can  use  that  letter  in  our  calculations, 
in  the  same  manner  as  if  it  were  the  true  answer. 

We  will  explain  by  a  few  examples. 

Example  1.  I  have  27  apples  which  I  wish  to  give  to  two 
children,  John  and  Mary,  in  such  a  manner  that  Mary  shall 
have  twice  as  many  as  John.  The  following  will  be  our 
method  of  rinding  how  many  each  shall  have. 

I  wish  to  give  to  John  a  certain  number,  and  to  Mary 
twice  that  number.  Therefore,  instead  of  saying,  I  will 
give  the  27  apples  to  John  and  Mary,  I  may  say,  I  will  give 
away  a  certain  number,  and  then  again  twice  that  number. 

Now,  a  certain  number,  and  then  twice  that  number, 
make,  in  all,  three  times  that  certain  number. 

But,  I  give  away  27  apples ;  and  hence  I  know  that 

Three  times  that  certain  number  is  just  as  much  as  27. 

Therefore,  that  number  itself  is  \  of  27,  which  is  9. 

Hence,  John  has  9  apples ;  and  Mary  has  twice  as  many, 
which  is  18. 

11 


12  ALGEBRA. 

§  3.  If,  in  the  operation  which  we  have  just  per- 
formed, we  use  the  letter  iV,  instead  of  the  words,  a  cer- 
tain number,  our  work  will  be  somewhat  abridged.     Thus : 

John  will  have  a  certain  number,  which  is  N. 

Mary  will  have  twice  as  much,  which  is  twice  N. 

And  both  together  will  have  three  times  N. 

But,  in  the  question,   we  said  both  shall  have  27 

Therefore,  three  times  N  is  the  same  as  27:  and  once 
N  is  j  of  27,  which  is  9;   which  is  John's  share. 

§  4.  About  the  year  1554,  Stifelius,  a  German,  in- 
troduced the  sign  -f  for  the  words  added  to;  and  called 
it  plus.  Ever  since  that  time,  -f  has  been  used  to  sig- 
nify that  the  quantity  after  it  has  been  added  to  the 
quantity  before  it.  Thus,  2  -f  6,  is  read  2  plus  6,  and 
signifies  2  with  6  added  to  it. 

Example  2.  A  brother  told  his  sister  that  he  was  4  years 
older  than  she;  and  that  his  age  and  her  age,  when  put 
together,  made  18  years.     What  was  the  age  of  each? 

By  the  statement,  his  age  was  the  same  as  her  age  with 
4  years  put  with  it ;  because  it  was  4  years  more  than  hers. 
Therefore  we  will  represent  the  sister's  age  by  A 

And  his  age,  which  is  4  years  more,  will  be  A  +  4 

Both  ages  together,  will  be  A  and  A  +  4,  or  A  +  A  +  4 

Which,  because  the  A's  can  be  put  together,  is    twice  A  -f-  4 
But  both  their  ages  together  made  18 

Therefore  twice  A  -f  4  is  the  same  as  18, 

And  twice  A  +  3  is  the  same  as  17, 

And  twice  A  +  2  is  the  same  as  16, 

And  twice  A  -J-  1  is  the  same  as  15, 

And  twice  A  alone  is  the  same  as  14, 

And  A,  or  the  sister's  age,  is  |  of  14,  which  is  7. 

§5.  About  the  year  1550,  John  Scheubelius,  a  Ger- 
man, introduced  the  following  practice.  Instead  of  writ- 
ing 2  times  A,  or  3  times  A,  or  4  times  A,  &c,  he 
wrote  2  A,  3  A,  4  A,  &c. 


PRELIMINARY    REMARKS.  18 

£xample  3.  Two  men,  A  and  B,  owe  me  $270 ;  and  B's 

cusbt  is  twice  as  much  as  A's.     How  much  does  each  of  them 
owe  me  ? 

We  will  represent  A's  debt  by  A. 

Then,  B's  debt  will  be  twice  as  much,  or  2^. 

And  both  debts  put  together,  make  A  +  2  A. 

But  both,  when  put  together,  are  equal  to  $270. 

Therefore,  A  -f-  2  A  is  the  same  as  $270. 

Putting  the  A's  together,  3  A  is  the  same  as  $270. 

Then,  one  A  is  \  of  $270,  which  is  $90. 

A's  debt  is  $90;  and  B's  debt  is  twice  as  much,  or    $180. 

§6.  In  1557,  Dr.  Recorde,  an  Englishman,  introduced 
the  sign  =,  which  we  call  equals.  It  is  used  instead  of 
the  expression,  is  the  same  as,  or,  is  as  much  as,  or,  is 
equal  to.     Thus,  2-j-6  =  8,  is  read,  2  plus  6  equals  8. 

Example  4.  B's  age  is  three  times  A's ;  and  C's 
age  is  double  of  B's ;  and  the  sum  of  all  their  ages  is 
70  years.     What  is  the  age  of  each  ? 

Let  us  represent  A's  age  by  the  capital  A, 

Then,  B's  age  will  be  three  times  as  much;  that  is,      3 A. 
And  C's  age  will  be  twice  3  A,  which  is  6  A. 

Then,  all  their  ages  put  together  will  be      A  +  3  A  -f  6  A. 
But  all  their  ages  put  together  is  70  years. 

Therefore,         A  +  3A  +  QA  =  70. 

Putting  them's  together,  10 ,#  =  70. 

Dividing  by  10,  A  =  7,  which  is  A's  age. 

A's  age  =7  years. 

B's  is  three  times  as  much;  or  21  years. 

C's  is  twice  B's  ;  or,  42  years. 

Proof,  "TO" 

§  7.  We  very  often  wish  to  state  that  we  have  made  a 
quantity  to  be  less.  This  was  formerly  done  by  using  the 
word  minus.  Thus,  if  I  wished  to  say  that  your  age  is  6 
years  less  than  mine ;  I  may  say,  Your  age  is  equal  to  mine 

2 


14  ALGEBRA. 

when  6  years  are  subtracted  from  mine ;  or,  in  fewer  words, 
your  age  is  equal  to  my  age,  minus  6  years. 

§  8.  Stifelius  introduced  the  sign  —  for  minus;  so  that 
20  —  6,  is  read,  20  minus  6;  and  signifies,  20  with  6  sub- 
tracted from  it.     Thus,  20  —  6  =  14. 

i 

Example  5.  At  a  certain  election,  548  persons  voted ;  and 
the  successful  candidate  had  a  majority  of  130  votes.  How 
many  voted  for  each  ? 

Let  us  represent  the  number  of  votes  received  by  the  sue- 
cessful  candidate,  by  the  letter  A.  Then,  as  the  unsuccess- 
ful candidate  received  130  votes  less,  his  number  may  be 
represented  by  A —  130. 

Both  candidates  have  A  +  A  —  130 

Then,  .#  +  .#-.130  =  548 

Putting  the  A's  together,  2  A  —  130  =  548,  which 

means,  2  A,  after  130  is  subtracted  from  it,  is  equal  to  548 
Now,  if  we  had  a  barrel  of  water,  minus  a  quart,  and  wanted 
a  full  barrel,  we  should  add  the  quart.     In  the  same  manner . 
as  we  have  2  A  minus  130,  we  must  add  the  130  to  get  th< 
complete  2  A. 

Therefore,  adding  the  130  to  both  sides,     2  A  =  678 

Dividing  by  2,  A  =  339. 

The  successful  candidate  had  339. 

The  unsuccessful  one,  130  less  =  209. 

Proof,  548. 

§  9.  Des  Cartes,*  a  Frenchman,  who  wrote  about  1637, 
used  the  last  letters  of  the  alphabet;  namely,  x,  y,  z,  w,  &c, 
to  denote  the  unknown  quantities.  And  this  is  now  the 
practice  of  all  mathematicians. 

Example  6.  It  is  required  to  divide  $300  among  A,  B, 
and  C  ;  so  that  B  may  have  twice  as  much  as  A,  and  C  may 
hi*ve  as  much  as  A  and  B  together. 

•  Pronounced  Da-Cart. 


PRELIMINARY    REMARKS.  15 

Let  us  represent  A's  share  by  x.     Then,  B's  share  will  be 

2  a*;  and  C  will  have  as  much  as  both  put  together,  which  is 

3  x.     Then, 

x  +  2x  +  3x  =  300. 
Putting  the  x's  together,  6x  =  300. 

Dividing  by  6,  x=    50. 

Therefore,  A's  share  is  $50. 

B's  is  twice  as  much,  or  $100. 

C's  =  as  much  as  both,  or         $150. 
Proof,  $300. 

From  what  we  have  shown  thus  far,  it  may  be  seen  that 
Algebra  is  a  kind  of  language,  made  up  with  letters  and  signs 
in  such  a  manner  that  all  the  reasonings  which  are  necessary 
for  the  solution  of  a  question,  may  be  contained  in  a  very 
small  space,  and  be  perceived  with  great  facility. 

§  10.  As  we  sometimes  wish  to  speak  of  particular  parts 
of  our  calculations,  mathematicians  have  given  the  name  term 
to  any  quantity  that  is  separated  from  others  by  one  of  the 
signs  -j-  or  — .  Thus,  in  the  last  example,  and  first  line  of 
the  operation,  the  first  x  is  the  first  term,  the  2  x  is  the  next 
term,  and  the  3a"  is  the  next  term;  and  the  300  is  the  last 
term. 

§11.  When  a  figure  is  put  before  a  letter,  to  denote  how 
many  times  we  take  the  quantity  which  that  letter  stands  for, 
the  figure  is  called  a  co-efficient.*  Thus,  in  the  term  2  a?,  2 
is  a  co-efficient  of  a:;  in  the  term  3*#,  3  is  a  co-efficient  of  .#. 

§  12.  It  must  also  be  understood,  that  a  letter  without  any 
number  before  it,  has  1  for  its  co-efficient.  Thus,  x  repre- 
sents la:1;  a  is  the  same  as  la,  &c.  The  1  is  omitted  be- 
cause it  is  plainly  to  be  understood. 

§  13.  When  any  algebraic  term  stands  by  itself,  it  is  called 
a  simple  quantity.  Thus,  i  is  a  simple  quantity;  2a*  is  a 
simple  quantity;   12  is  a  simple  quantity,  &c. 

*  Thin  name  was  given  by  Franciscus  Vieta,  about  1573. 


16  ALGEBRA. 

§  14.  Quantities  that  consist  of  more  than  one  term, 
are  called  compound  quantities.  Thus,  a  +  b  is  a  com 
pound  quantity;  so  is  also  a  —  6,  and  a?-f  7,  and  x —  7 
and  a  +  x — 7  —  6,  &c. 

§  15.  When  any  algebraic  quantity  begins  with  the  sign 
4-,  it  is  called  a  positive  quantity;    as,  -fa,  -f  3  a. 

§  16.  When  any  algebraic  quantity  begins  with  the  sign 
— ,  it  is  called  a  negative  quantity  ;  as,  — 6,  — 5  a. 

§  17.  In  algebra,  the  perfect  representation  of  any  simple 
quantity  requires  both  the  specified  sum,  and  either  the  sign 
-f ,  or  the  sign  — ;  as,  -f  5,  —40,   -f  x,  — 3x. 

§  18.  But,  when  a  positive  quantity  stands  by  itself,  or 
when  it  is  the  first  term  of  a  compound  quantity,  the  sign 
that  belongs  to  it  is  generally  omitted  on  paper,  and  also  in 
our  reading  ;  as,  J*,  2,  a  -f  6,  x — y. 

§  19.  Therefore,  when  a  simple  quantity,  or  the  first  term 
of  a  compound  quantity,  does  not  begin  with  a  sign,  we  say 
that  the  sign  -f  is  understood.  That  is,  we  think  of  the 
quantity  the  same  as  if  -f  was  before  it. 

§  20.  In  reading  compound  quantities,  the  pupil  must  be 
careful  to  join  the  sign  to  the  term  that  is  immediately  after 
it.     Thus,  the  following  quantity  must  be  read  a;  plus  b; 
plus  8  ;  minus  4 ;  plus  2  a  ;  plus  6 ;  minus  36: 
a  +  &  +  8  —  4  +  2a-f  6  —  3  6. 

§  21.  When  a  quantity  is  expressed  by  figures,  it  is  called 
a  numeral  quantity.  When  it  is  represented  by  a  letter,  it  is 
called  a  literal  quantity. 


SIMPLE    EQUATIONS. 

§  22.'  The  most  general  and  useful  application  of  algebra, 
is  that  which  investigates  the  values  of  unknown  quantities 
by  means  of  equations. 


SIMPLE    EQUATIONS.  17 

§  23.  An  equation  is  an  expression  which  declares  one 
quantity  to  be  equal  to  another  quantity,  by  means  of  the  sign 
=  being  placed  between  them. 

Thus,  5  -f  3  =  8,  is  an  equation,  denoting  that  5  with  3 
added  to  it,  equals  8.  Also,  4 — 1=3,  and  3  +  2  — 1=4, 
and  8 — 2  =  5  +  1,  are  equations,  each  denoting  that  the 
quantity  on  one  side  of  the  =,  is  equal  to  that  on  the 
other  side. 

§  24.  The  whole  quantity  on  the  left  of  =  is  called  the 
first  member  of  the  equation;  and  all  on  the  right  of  =  is 
called  the  last  member  of  the  equation. 

In  order  to  be  a  member  of  the  equation,  it  is  of  no  import- 
ance whether  the  quantity  is  simple  or  compound.  Thus,  in 
the  equation  x  =  4  -f-  a — b — 18,  a?  is  the  first  member,  and 
4  -f  a — b  — 18  is  the  last  member.  And  in  this  case,  the  first 
member  represents  just  as  great  a  quantity  as  the  last. 

§  25.  In  order  that  an  equation  may  be  such  that  it  will 
enable  us  to  find  the  value  of  an  unknown  quantity  by  it,  it 
must  contain  some  quantity  that  is  already  known.  And  then 
we  can  find  the  value  of  the  unknown  quantity,  by  making 
it  stand  by  itself  on  one  side  of  =,  and  all  the  known  quan 
tities  on  the  other  side. 

§  26.  But  it  is  generally  the  case  that  the  known  and  un 
known  quantities  are  mingled  together.  And  then  we  are  to 
alter  the  arrangement  of  the  terms,  so  as  to  bring  the  unknown 
quantity  by  itself;  and  this  must  be  done  without  destroying 
the  equality  of  the  sides.  This  operation  is  called  solving  or 
reducing  the  equation. 

§  27.  When  the  known  quantities  are  represented  by  num- 
bers, the  equation  is  called  a  numerical  equation.  It  is  to 
this  kind  of  equations  that  the  first  part  of  this  treatise  relates. 

§  28.  In  solving  equations,  the  different  arrangement  of  the 
terms  is  brought  about  by  addition,  subtraction,  multiplication, 
division,  &c,  as  the  case  may  require. 
B  2* 


18  ALGEBRA.  [SECTION  I. 

SECTION  I. 
SIMPLE   ALGEBRAICAL  .OPERATIONS. 

I.      ADDITION    AND    SUBTRACTION    OF    SIMPLE    QUANTITIES. 

§  29.  In  algebra,  simple  quantities  are  added  by  writing 
them  down  one  after  another;  being  careful  to  retain  the  signs 
that  are  expressed  or  understood.  Thus,  we  add  9  to  x,  by- 
writing  them  x  +  9 ;  because,  when  those  quantities  were 
alone,  -f  was  understood  before  9.  We  add  — 9  to  #,  by- 
writing  them  x — 9. 

89"  The  pupil  must  understand  that  x  stands  for  some 
number;  but  it  is  often  the  case  that  we  do  not  know  what 
that  number  is.  For  it  may  stand  sometimes  for  one  number, 
and  sometimes  for  another. 

§  30.  It  will  readily  be  seen,  that  it  is  of  no  consequence 
which  quantity  is  put  first ;  for  9  -j-  7  is  the  same  amount  as 
7-f  9. 

§31.  One  positive  simple  quantity  is  subtracted  from 
another  simple  quantity,  by  writing  down  both  quantities 
one  after  another,  and  changing  from  +  to  —  the  sign  be- 
fore the  quantity  which  we  subtract.  Thus,  we  subtract  4 
from  x,  by  writing  them  x — 4;  which  is  read  x  minus  4; 
and  not  4  from  x. 

Questions.  What  is  algebra?  See  §  1.  In  what  manner  do  we 
use  letters  in  our  calculations'?  2.  What  is  said  of  the  sign  -+-  ?  4. 
—  ?  7.  =  ?  6.  What  does  term  signify  in  algebra?  10.  What  is 
a  simple  quantity  ?  13.  How  are  simple  quantities  added  in  alge- 
bra? 29.  What  is  requisite  for  the  perfect  representation  of  an 
algebraic  term?  17.  Are  all  terms  thus  represented?  What  is 
understood  in  such  cases?  is  the  sign  —  ever  understood?  In 
addition,  which  quantity  must  be  put  first?  30.  What  particular 
number  does  a  letter  stand  for  in  algebra?  Why  do  we  use  letters 
in  algebra?  2.  How  is  one  simple  quantity  subtracted  from  an- 
other? 31.     Which  quantity  must  have  —  before  it? 


§  36. J  SIMPLE    ALGEBRAICAL    OPERATIONS.  19 

§  32.  When  two  simple  quantities  have  been  added  toge- 
ther, or  one  simple  quantity  subtracted  from  another,  the 
answer  will  consist  of  more  than  one  term,  and  become  a 
compound  quantity. 

§33.  It  sometimes  happens,  that  in  those  compound 
quantities^  which  are  made  by  adding  or  subtracting,  there  are 
two  or  more  lerms  of  the  same  kind ;  that  is,  such  terms  as 
do  not  differ  at  all,  or  differ  only  in  their  numeral  co-effi- 
cients;  such  as  a?  +  2a?;  4a— 2a;  5a?  — 3a?  +  a?.  Such 
quantities  are  called  similar  quantities. 

§  34.  When,  in  any  compound  quantity,  there  are  two  or 
more  terms  of  the  same  kind,  they  may  be  united  by  per- 
forming arithmetically  the  operation  which  is  expressed  by 
the  signs  belonging  to  them.  Thus,  a?  +  2a?  is  united  into 
3a?;  5a  —  3a  is  united  into  2a;  5a? — 3a?  -fa?  is  united  into  3a?. 

§  35.  Numeral  quantities  may  be  united  in  the  same  man- 
ner. Thus,  4  +  3  may  be  united  into  7 ;  9  —  4  may  be  united 
into  5  ;  6  —  2  +  5  may  be  united  into  9. 

Examples.  The  pupil  may  unite  the  following  quantities : 


1.  a  +  3a+5a. 

2.  x +  5a? +  x. 

3.  10a? — 4a? — a?. 

4.  a+lla  -2a. 


5.  b  +  6b—3b. 

6.  la — 4a-fa. 

7.  4  +  8—3. 

8.  5  —  2+7. 


9.  3a?+2.r+8— 4. 

10.  10a—  3a+7— 2. 

11.  6a+17— 3a— 7. 

12.  14  +  5a?—  3a;—  5. 


2.    GENERAL    RULE    FOR    UNITING    TERMS. 

§  36.  1st.  When  similar  quantities  have  the  same  sign, 
select  one  kind,  and  find  the  sum  of  the  numeral  co-efficients ; 


Questions.  What  is  a  compound  quantity  ?  14.  What  operation 
will  produce  compound  quantities?  What  are  similar  quantities'? 
What  is  a  co-efficient]  11.  Have  all  terms  a  co-efficient  express- 
ed? 12.  What  may  be  done  with  similar  quantities?  When  a 
sign  is  between  two  quantities,  to  which  quantity  does  it  be- 
long? 20.  How  does  a  literal  quantity  differ  from  a  numeral  ?  21. 
What  is  a  positive  quantity  ?  15.  What  is  a  negative  quantity  ?  16 
What  is  the  first  part  of  the  rule  for  uniting  terms  ?  What  is  the 
second  part  of  the  rule  for  uniting  terms  ? 


20  ALGEBRA.  ^SECTION  I. 

and,  preserving  the  sign,  annex  the  common  letter.     Then 
select  another  kind,  and  proceed  in  the  same  manner  with  that. 

EXAMPLES. 

1.  Unite  the  following  terms.     5a?  — 3?/ -f- 4a: — ly. 

Ans.  The  x's,  when  united,  equal  -f-  9a? »  and  the  y's, 
when  united,  equal  — l(h/.     Ans.  9x — lOy. 

2.  2x  +  3y  +  Sx  -f  Ay  +  Ix  -f  y  -f  x  -f  9y.      Ans.    13a? 

*$&sT  As  fast  as  the  pupil  adds  the  co-efficients,  he  should 
slightly  cross  the  term,  so  that  he  may  know  when  all  the 
terms  have  been  united.     Thus  : 

2p+3y  +  3$+4y+7v+y+.V+9y=lZx+l7y. 

3.  4a— 3x+2a— x+a  +  3a— Sx  —  x.     Ans.   10a  —  8a:. 

4.  — 2y+7— a?— y— 3a?  +  ll— 2t/.     Ans.   18— 5y— 4a\ 

5.  —  2a— a?+3— a— 3a:  +  4— 4a.     Ans.  — 7a— 4a?+7. 

§  36.  2d.  When  similar  quantities  have  different  signs, 
select  one  kind,  and  add  into  one  sum  all  the  p  isitive  co-effi- 
cients that  belong  to  them.  Then  add  into  another  sum  all 
the  negative  co-efficients  that  belong  to  them.  Subtract  the 
less  sum  from  the  greater,  and  prefix  the  sign  of  the  greater 
to  the  difference ;  annex  the  common  letter.  Then  select 
another  kind,  and  proceed  in  the  same  manner. 

EXAMPLES. 

Unite  the  terms  in  the  following. 

6.  3a  +  46  +  26— 3c  +  3c  +  a— 6— 3b. 

Ans.  The  a's,  when  united,  equal  -\-4a;  the  6's,  when 
united,  equal  +66 — 4b,  which  equals  4-26.  Then,  — 3c  +  3c 
balance  each  other,  so  as  to  be  equal  to  0.  The  answer  is 
4a +26. 

7.  2a-|-6+4a  +  46  +  8-|-6— 3a— 36— 4.  Ans.  3a +26 +  10. 

8.  3x+y+z  +  3x  +  3y—4x—2y—z.     Ans.  2x  +  2y. 

9.  18  +  8a  +  4a:  +  7.r  +  5a— 6x— 7a.     Ans.   18  +  6a  +  5a\ 
10.  4y+7z — 4z — y  +  2z  +  2y — z.     Ans.  5y  +  4z. 


§  36.]  UNITING   TERMS.  21 

Note. — It  sometimes  happens  that  the  negative  quantity  is 
greater  than  the  positive  quantity.  In  such  cases,  the  differ- 
ence will  have  the  sign  — . 

11.  la— 3a+4z+a— 2z— 4a—  6a.     Ans.  —  5a+2*. 

12.  a+6  +  3a— c+4a— 3c  +  6.     Ans.  8a+26— 4c. 

13.  5x  +  5y +  3x—2y.     Ans.  8x+3y. 

14.  6+a+3a— 56+4cc— a+3a— x.     Ans.  6a— 46  +  3.T 

15.  4a— 56— \0  +  2y— x  +  4.     Ans.  4a—  56+ 2y— x— 6. 

16.  2a?+  3?/— 3x+5y— x— z.     Ans.  —  2x+8y— z.~ 

17.  #+z+;c— z+#+3?/.     Ans.  3x+3y. 

18.  3a— 5  —  46  +  6— 2a— 3  +  66.     Ans.  a+26— 2. 

19.  4;r+4  +  5?/+6— 2+3a?— 2y.     Ans.  7a?+3y  +  8. 

20.  4c  +  3a— 6  — 3a+c— 6a— 6.     Ans.  5c— 2a— 26. 

21.  26— c+3a  — 6+4c  +  2tf— 6  — 5a\  -=         U^V  «  * 

22.  a— 6  +  c+a*+3a+6  — 2c— 4a+10+a. 

23.  a+6  +  36+#— 7+4:r— 6+3a— y+2. 

24.  a?— 4+10—  7.r  +  a— 6—  10  +  2a—  1.  v 

25.  3a— 6+4c+2:c— c  +  4a+5c+76. 

26.  10— a  +  6  —  6  —  a?+7  +  36+2a— 40. 

27.  8a— 16  — z  +  91+2y  — 87— 3z  +  14. 

28.  29  +  46+27+y— 32— 43+y.  r 

29.  73+4x— 36  — 3^—41+7^— y+2.r.     . 

30.  a— 6— a-6-67— 42  +  7a+36./ 

31.  4X_37/+2a— lx+6y  —  7a+5x— 7y.  h2^ 

32.  4x—3x-\-4-\-x— 2x— 5  +  l  +  3a?— 5a?.  "-P-vX- 

33.  2Z+1/+9— a:— i/— 9  +  3#+10a?y.  It   ^fc-^'^Z 

34.  5a+36— 4c+2a— 56  +  6c+a— 46— 2c.  9 VL.-C4 

35.  3a+;r+10— 5a+2x— 15— 4a— 10z+21  -  i^"^  +lv 

36.  5a— 36+4a— 76+7a+36— 5a— 96.  H.     ' 

37.  6a+2;r—  10+2a— 3a?+10— 2y— 3a+2x. 


22  ALGEBRA.  [SECTION  *• 

III.     ADDITION    OF    COMPOUND    QUANTITIES. 

§  37.  When  two  or  more  expressions  that  consist  of  seve- 
ral terms,  are  to  be  added  together,  the  operation  is  represented 
by  writing  them  one  after  another ;  taking  care,  in  all  except 
the  first,  to  insert  the  signs  that  are  understood.  Thus,  a — x 
is  added  to  y-\-l  in  the  following  manner: 

a — a? -f  ?/4-7,  or  y-\-l-\-a — x. 
For  to  a —  x  we  add  first  y,  and  then  7  ;  or  to  y-f  7  we  add 
a,  and  then  because  we  ought  to  have  added  a  —  a?,  we  see 
that  we  have  added  x  too  much,  and  therefore  subtract  it. 

§38.  The  above  example  shows  that  it  is  of  no  conse- 
quence in  what  order  we  write  the  terms.  Their  place  may- 
be changed  at  pleasure,  provided  their  signs  be  preserved. 

EXAMPLES. 

1.  Add  the  following  compound  quantities.  2«  —  8a?, 
x  —  3a,   — 4a —  2a?,   4x —  a.     Ans.  — 6a  —  5a?. 

2.  Add  2  —  a? +  4?/,  3  -f  3a?— y,  —  30  —  x  —  2y,  and 
1  —  2a? + 3y  —  1  Oz  together.     Ans.  —  24 — a?  -f  4y  —  1  Oz. 

3.  Add  3x+5y— 6z,  — 2a?— 8y— 9z,  20a? -f  2y— 3z,  and 
x—y+z—4  together.     Ans.  22a?— 2y— 17 z— 4. 

4.  Add  3— 2y+z,  4y— 2z-f5,  2—z—y,  and  2z— y— 10. 
Ans.  Nothing. 

5.  Add  3a?— 6,  4a?+2,  64-3a?,  7— 2a?,  ar-f-1.  £->+  /  J 

6.  Add  4a?— a,   3a— a?,  4a?+a,  7a— 3a?.    /  p  &-  *+  */" 

7.  Add2+2*4-3z,  3z  +  8,   5— 2z,  4z.  If  +? 

8.  Add  4— a?,  —a?— 3,   6+2a?,  a?+2a?—  1. 

9.  Adda?— 21,  3+4a?+a?,  2a?+3a?+36.  /  /  %~t  '  ^ 

10.  Add  a?,  a?— 3,  a?+a?— 3,  a?+2a?— 6. 

11.  Add  2a?— 3z-4,  4z— 3a?,   5a?— 5,   3— 10a?+3. -  kT^ 

12.  Add3?/-4,    6-3a?,   7x—5y,   x—y—2,   3y  +  5x—3. 

Questions. — How  are  compound  quantities  added  ?  Why  is  the 
sign  —  preserved  in  addition  1   Which  expression  must  be  put  first! 


§  43. J  MULTIPLICATION    AND    DIVISION.  23 

IV.    MULTIPLICATION    AND    DIVISION    OF    SIMPLE    QUANTITIES. 

"§39.  Were  we  to  add  a  to  itself  four  times,  we  should 
write  the  sum  thus,  a  +  a  +  a+a-,  which,  when  united,  be- 
comes 4a.     Whence  we  have  the 

Rule. — Any  literal  quantity  is  multiplied  by  a  number, 
by  putting  that  number  before  it  as  a  co-efficient;  as,  7  times 
x  is  7a?.   6  times  a  is  6a. 

§  40.  If  the  quantity  to  be  multiplied  has  already  a  co- 
efficient, that  co-efficient  only  is  to  be  multiplied.  Thus,  3x 
taken  four  times  is  3a?-f 3a?-j-3a?-f 3a?,  which  when  united 
=  12a?.  The  co-efficient  is  multiplied  by  4.  Thus,  4  times 
3a?  =  12a\ 

§  41.  It  is  evident  that  if  2  times  3a?  is  6a?,  then  one-half 
of  6a?  is  3a?.  Whence  we  learn  that  a  quantity  with  a  nu- 
meral co-efficient,  may  be  divided,  by  merely  dividing  that 
co-efficient.  Thus,  8a?  divided  by  2  =4a?.  12a?  divided  by 
4  ==  3a?,  &c. 

§  42.  In  1661,  Rev.  William  Oughtred  of  England,  pub- 
lished a  work,  in  which  he  introduced  the  sign  x  to  represent 
multiplication.  Thus,  4x3=12,  is  read  4  multiplied  by  3 
equals  12 ;  or,  4  into  3  equals  12. 

§43.  In  1668,  Mr.  Brancker  invented  the  sign  -r-  for  di- 
vision. The  sign  is  always  put  before  the  divisor;  as, 
20  h-  4  =  5 ;  read  20  divided  by  4  equals  5 ;  or,  20  by  4, 
equals  5. 

But  in  algebra,  division  is  frequently  performed  by  writ- 
ing the  divisor  under  the  dividend,  so  as  to  make  a  fraction ; 
thus,  20  divided  by  4,  is  \°.     See  page  48. 

Questions. — How  do  we  multiply  a  literal  quantity?  Multiply  x 
by  3,  4,  5,  6,  &c.  to  20.  How  do  we  multiply  a  quantity  that  has 
already  a  co-efficient?  Multiply  2x  by  3,  4,  5,  &c.  to  12.  Multiply 
3x  by  the  same  numbers.  Multiply  ix  by  the  same.  How  do  we 
divide  a  quantity  that  has  a  numeral  co-efficient]  Describe  the  sign 
for  multiplication.     Describe  the  sign  for  division. 


24  ALGEBRA.  [SECTION  1 

EQUATIONS.— SECTION  1. 
Equations  which  are  solved  by  merely  uniting  terms. 

In  each  of  the  following  equations,  ihe  object  is  to  find  the 
value  of  x. 

Example  1.     a?+2a?  =  45— 15. 
Uniting  terms,  Sx  =  30. 

Now,  as  we  have  found  that  three  x's  =  30,  it  is  evident 
that  one  x  will  be  one-third  of  30. 

Therefore,  dividing  by  3,  x  =  10. 

2.  Find  x  in  8a:  —  4a?  —  #  =  7+26+51  — 15 

Uniting  terms,  3a?  =  69 
Dividing  by  3,    a?  =  23. 

3.  Find  a:  in  10a?— 5a?+4a:  =  56+75  +  32  —  1 

Uniting  terms,  9a:  =  162 
Dividing  by  9,    x  =  18. 

4.  Find  x  in  a?+2a:+3a:+4a?  =  12+35+74— 11 

Uniting  terms,  10a?  =110 
Dividing  by  10,    a?  =  ll. 

5.  8a:— 3a?+2a?  =  46  +  54  +  37— 4.  Ans.  a?=19. 

6.  4a?— 3a? + 4a?  =  29—36+48+ 14.  Ans.  a?  =  11. 

7.  6a?— 8a-+ 14a?  =12  +  36  + 14+22.  Ans.  a?  =  7. 

8.  5a?+4a:+3a'  =  49  +  14'+22  +  ll.  Ans.  a?  =  8. 

9.  7a?+a*=14— 22— 11+41— 6.  Ans.  a?  =  2. 

10.  4a:— 2a?  =  96— 7  +  8— 15— 10.  Ans.  a?  =  36. 

11.  5a—  a?  =  2  +  3— 15— 10+72.  Ans.  a?  =  13. 

12.  6a?  =  7  +  4+72— 51  — 16— 10.  Ans.  a?=l. 

Questions.  What  is  the  most  useful  application  of  algebra  ?  §  22. 
What  is  an  equation?  What  is  the  part  on  the  left  of  =  called"? 
What  the  part  on  the  right1?  What  if  there  are  more  terms  on  one 
side  than  on  the  other?  What  is  a  numerical  equation?  How  do 
we  find  the  value  of  the  unknown  quantity?  What  is  this  opera- 
tion called  ? 


§  47.]  EQUATIONS    SOLVED    BY    UNITING    TERMS.  25 

13.  Sx— 7x+5x— 4:X+3x=27— 12.  Ans.  x=3. 

14.  5#— 4a?-f2a?— 3#-fa:=39— 13.  Ans.  a?=26 

15.  17a:— 4a? — 2a;— 5a?— x =57— 32.  Ans.  #=5. 

16.  14a?— 35a?-f29a:+47a?-f-a:=504.  Ans.  a?=9. 

EQUATIONS    FORMED    FROM    ARITHMETICAL    QUESTIONS. 

§  44.  An  algebraic  problem  is  a  proposition  which  sup 
poses  that  an  unknown  quantity  has  certain  relations  with 
other  quantities  that  are  known ;  and  requires  the  disco- 
very of  some  arithmetical  operation  by  which  the  unknown 
quantity  may  be  ascertained. 

§  45.  Problems  properly  belong  to  literal  algebra ;  but 
because  questions  in  numeral  algebra  are  solved  in  a  simi- 
lar manner,  they  also  are  called  problems  in  this  work. 

§  46.  In  the  solution  of  problems,  the  first  thing  to  be 
done,  is  to  represent  the  required  answer  by  x,  y,  or  some 
other  final  letters  of  the  alphabet.  Then,  make  in  alge- 
braical language,  with  its  explanation,  a  statement  of  each 
of  the  conditions  in  the  question,  in  the  same  manner  as  if 
the  letter  were  the  true  answer,  and  you  were  required  to 
prove  it. 

§  47-  When  the  question  has  been  fairly  stated,  it  will 
be  found  that  two  different  algebraical  expressions  have  the 
same  explanation.  These  two  expressions  must  be  put  in 
the  same  line,  with  the  sign  =  between  them,  so  as  to  form 
an  equation.  And  then,  by  reducing  the  equation,  the 
required  result  will  be  found.     See  §  25  and  26. 

I.  The  sum  of  $660  was  subscribed  for  a  certain  pur 
pose,  by  two  persons,  A  and  B ;  of  which,  B  gave  twice 
as  much  as  A.     What  did  each  of  them  subscribe  ? 

Questions.  What  is  an  algebraic  problem  ?  What  is  the  first  step 
in  the  solution  of  problems?  How  must  you  continue  the  statement? 
How  is  it  known  when  the  statement  is  perfect?  What  is  the  second 
part  of  the  solution  ? 

3 


26  ALGEBRA.  ^SECTION  I. 

Stating  the  question,  x  dollars  =  what  A  gave. 

2x  dollars  =  what  B  gave. 
X  -x-2x  dollars  =  what  both  gave. 
And  also,  660  dollars  =  what  both  gave. 

Therefore,  putting  the  question  into  an  equation, 
x  +2  x  =  660 
Uniting  terms,  3x  =  660 

Dividing  by  3,  x  =  220  A's  share. 

2x  =  440  B's  share. 
660  proof. 

2.  Three  persons  in  partnership,  put  into  the  stock  $4800  ; 
of  which,  A  put  in  a  certain  sum,  B  twice  as  much,  and  C  as 
much  as  A  and  B  both.     What  did  each  man  put  in  ? 

Stating  the  question,  x  =  A's  share* 

2x  =  B's  share. 

x  +  2x  =  C's  share. 

Adding  all  three  shares,      x  -f  2x  -f-  x  +  2x  «■  the  whole. 

4800  =  the  whole. 
Therefore,  forming  the  equation, 

x  -f  2x  -f  x  -f  2x  —  4800 
Uniting  terms,  Gx  =  4800 

Dividing  by  6,  x  =    800  A's  share. 

2x  =  1600  B's  share. 
800  -f  1600  =  2400  C's  share. 
4800  proof. 

E^~  In  all  the  succeeding  problems,  the  learner  should 
prove  his  answers. 

3.  A  person  told  his  friend  that  he  gave  108  dollars  for  his 
horse  and  saddle  ;  and  that  the  horse  cost  8  times  as  much  as 
the  saddle.     What  was  the  cost  of  each  1 

Stating  the  question,  x  =  price  of  the  saddle. 

Sx  =  "  horse. 

x  +  8x=  "  both. 

108  =  "  do. 

Forming  the  equation     x  -f  Sx  =  108 


§  47.]  EQUATIONS   SOLVED    BY   UNITING   TERMS.  27 

Uniting  terms,  9x  =  108 

Dividing  by  9,  x  =  12  =  price  of  saddle. 

96  =  "  horse. 
It  is  advisable  for  the  pupil,  while  performing  his  sums,  to 
write  them  on  his  slate  in  a  manner  similar  to  the  three  ques- 
tions above;  beginning  the  statement  by  making  x  the  answer 
to  the  question,  and  throughout  the  operation  keeping  equals 
under  equals.  And  in  recitation,  the  whole  of  it  is  to  be 
recited. 

4.  A  father  once  said,  that  his  age  was  six  times  that  of 
his  son;  and  that  both  of  their  ages  put  together,  would 
amount  to  49  years.     What  was  the  age  of  each  ? 

Ans.  Son's  age  7  years ;  father's  42. 

5.  A  farmer  said  that  he  had  four  times  as  many  cows  as 
horses,  and  five  times  as  many  sheep  as  cows ;  and  that  the 
number  of  them  all  was  100.  How  many  had  he  of  each 
sort?  Ans.  4  horses  ;  16  cows;  and  80  sheep. 

6.  A  boy  told  his  sister  that  he  had  ten  times  as  many 
chestnuts  as  apples,  and  six  times  as  many  walnuts  as  chest- 
nuts. How  many  had  he  of  each  sort,  supposing  there  were 
639  in  all  ?     Ans.  9  apples  ;  90  chestnuts  ;  and  540  walnuts. 

7.  A  school-girl  said  that  she  had  120  pins  and  needles ; 
and  that  she  Had  seven  times  as  many  pins  as  needles.  How 
many  had  she  of  each  sort?     Ans.   15  needles,  and  105  pins. 

8.  A  teacher  said  that  her  school  consisted  of  64  scholars ; 
and  that  there  were  three  times  as  many  in  arithmetic  as  in 
algebra,  and  four  times  as  many  in  grammar  as  in  arithmetic. 
How  many  were  there  in  each  study? 

Ans.  4  in  algebra;   12  in  arithmetic  ;  and  48  in  grammar. 

9.  A  teacher  had  four  arithmeticians  who  performed  80 
sums  in  a  day.  The  second  did  as  many  as  the  first,  the 
third  twice  as  many,  and  the  fourth  as  much  as  all  the  other 
three.  How  many  did  each  perform  ?  Ans.  The  first  and 
second,  each  10 ;  the  third,  20 ;  and  the  fourth,  40. 


28  ALGEBRA.  ^SECTION  I. 

10.  A  person  said  that  he  was  $450  in  debt.  That  he 
owed  A  a  certain  sum,  B  twice  as  much,  and  C  twice  as  much 
as  to  A  and  B.     How  much  did  he  owe  each  ? 

Ans.  To  A  $50,  to  B  $100,  and  to  C  $300. 

11.  A  person  said  that  he  was  owing  to  A  a  certain  sum; 
to  B  four  times  as  much ;  and  to  C  eight  times  as  much ;  and 
to  D  six  times  as  much ;  so  that  $570  would  make  him  even 
with  the  world.     What  was  his  debt  to  A  ?  Ans.  $30. 

12.  A  boy  bought  some  oranges  and  some  lemons  for  54 
cents.  The  price  of  the  oranges  was  twice  the  price  of  the 
lemons.     How  much  money  did  he  spend  for  each  sort? 

Ans.  18  cents  for  lemons  ;  and  36  cents  for  oranges. 

13.  A  boy  bought  some  apples,  some  pears,  and  some 
peaches,  an  equal  number  of  each  sort,  for  72  cents.  The 
price  of  a  pear  was  twice  that  of  an  apple,  and  the  price  of  a 
peach  was  3  times  that  of  an  apple.  How  much  money  did 
he  give  for  each  kind? 

Ans.  12  cents  for  apples ;  24  for  pears  ;  36  for  peaches. 

14.  A  man  bought  3  sheep  and  2  cows  for  $60.  For  each 
cow,  he  gave  6  times  as  much  as  for  a  sheep.  How  much 
did  he  give  for  each  ? 

{y  If  x  =  price  of  a  sheep,  all  the  sheep  will  cost  three 
times  as  much,  or  3a\  In  the  same  manner  both  cows  will 
cost  twice  as  much  as  one  cow.  One  cow  will  cost  6a?,  and 
2  cows  will  cost  Ylx. 

Ans.  $4,  price  of  a  sheep  ;  and  $24  price  of  a  cow. 

15.  A  gentleman  hired  3  men  and  2  boys.  He  gave  five 
times  as  much  to  a  man  as  he  gave  to  a  boy ;  and  for  all  of 
ihem  he  gave  $6.80.     What  was  the  wages  of  each  ? 

Ans.  A  boy's  wages  was  40  cents,  and  a  man's  wages,  $2 

16.  Two  men,  who  are  560  miles  apart,  start  to  meet  each 
other.  One  goes  30,  and  the  other  goes  40  miles  a  day.  In 
how  many  days  will  they  meet? 

Each  will  travel  x  days.     The  first  will  go  x  times 


§  47.]  EQUATIONS    SOLVED    BY    UNITING    TERMS.  29 

30  miles,  and  the  second  will  go  x  times  40  miles ;  and  both 
together  will  go  the  whole  distance.  It  is  also  evident  that  x 
times  30  is  the  same  as  30  times  x ;  &c.  Ans.  8  days. 

17.  A  farmer  hired  three  labourers  for  $50.00 ;  giving  to 
the  first  $2.00  a  day,  to  the  second  $1.50,  and  to  the  third 
$1.00.  The  second  worked  three  times  as  many  days  as  the 
first ;  and  the  third  twice  as  many  days  as  the  second.  How 
many  days  did  each  work  ? 

Ans.  The  first,  4;  second,  12  ;  and  third,  24  days. 

18.  A  gentleman  bought  some  tea,  coffee,  and  sugar,  for 
$7.04 ;  giving  twice  as  much  a  pound  for  coffee  as  for  sugar, 
and  five  times  as  much  for  tea  as  for  coffee ;  and  there  were 
20  pounds  of  sugar,  12  pounds  of  coffee,  and  2  pounds  of  tea. 
What  was  the  price  of  each?  Ans.  11  cents  for  sugar;  22 
cents  for  coffee ;  and  110  cents  for  tea. 

19.  A  bookseller  sold  to  a  teacher  at  one  time  10  books, 
and  afterwards  15  more  at  the  same  rate.  Now  the  difference 
between  the  whole  sum  received  at  the  latter  time,  and  the 
whole  sum  received  at  the  former  time,  was  60  cents.  What 
was  the  price  of  each  book?  Ans.   12  cents. 

20.  In  fencing  a  side  of  a  field,  whose  length  was  450 
yards,  two  workmen  were  employed ;  one  of  whom  fenced  9 
yards  per  day,  and  the  other  6  yards  per  day.  How  many 
days  did  they  work  to  make  the  whole  fence  ?    Ans.  30  days. 


8» 


30  ALGEBRA.  ^SECTION  II. 

SECTION  II. 
TRANSPOSITION. 

TRANSPOSITION    BY    SUBTRACTION. 

§  48-  It  often  happens  that  in  the  first  member  of  the 
equation,  some  number  has  been  added  to  the  x*u  in  order  to 
make  them  equal  to  the  last  member.     Thus,  in  the  equation 

07+16  =  46 
we  see  that  16  has  been  added  to  #,  to  make  it  equal  to  46. 

§  49.  Now  if  x  with  16  added,  is  equal  to  46;  then  x 
alone  must  be  16  less  than  46;  that  is,  46 — 16.  So,  that  if 
we  find,  that  #+16  =  46,  we  may  know  that  a?  =  46— 16; 
or  what  is  the  same,  x  =  30. 

§  50.  But  this  may  be  proved  another  way.  It  is  very 
plain  that  if  we  subtract  a  quantity  from  one  member  of  an 
equation,  and  then  subtract  the  same  quantity  from  the  other 
member  of  the  equation ;  it  will  still  be  the  fact  that  the  two 
members  are  equal  to  one  another.  Thus,  a  half-dollar  =  50 
cents.  Subtract  2  cents  from  each  member.  Then  a  half- 
dollar  —  2  cents  =  50  cents  —  2  cents ;  for  each  of  them  is 
equal  to  48  cents. 

§  51.  Thus,  with  the  equation  that  we  had  above, 

#+16  =  46 
Subtracting  16  from  both  members,  a;+16 — 16  =  46 — 16. 

Now,  in  the  first  member  of  the  equation,  we  have  +16 
— 16,  which  is  of  no  value  at  ail,  for  +16  and  —16  balance 
each  other,  as  has  been  seen  in  Ex.  6  under  §  36. 
Therefore  the  equation  is  reduced  to  x  =  46 — 16 

Uniting  terms  in  the  last  member,  x  =  30. 


Questions.  Is  the  first  member  always  without  numeral  quanti- 
ties ?  When  x  with  another  number  added,  equals  a  certain  number 
what  is  x  itself  equal  to  ?     How  can  this  be  proved  t 


§  51-3  TRANSPOSITION    BY    SUBTRACTION.  31 

EQUATIONS.  — SECTION  2. 

1.  Suppose  a?+8+3a?  =  56;  what  is  the  value  of  a?? 
Uniting  terms,  4a?-f-8  =  56 
Subtracting  8  from  both  sides,  4a?+8— 8  =  56—8 
Which  is  the  same  as  4a?  =  56 — 8 
Uniting  terms  in  the  last  member,         4a?  =  48 
Dividing  by  4,  a?  =  12. 

2.  Suppose  2a?-fl4— a?— 7  =41+2—8;  to  find  a?. 
Uniting  terms,  a?-f-7  =  35 

Subtracting  7  from  £    a?-f  7— 7  =  35 — 7  or 

both  sides,        5  a?  =  35 — 7 

Uniting  terms,  x  =  28. 

3.  Given  a?-f5— 2a?— 3-f4a?  =  26,  to  find  x. 

Uniting  terms,  3a?+2  =  26 

Subtracting  2  from  both  sides,  3a?  =  24 
Dividing  by  3,  a?  =  8. 

4.  Given  5a?-f22— 2a?  =  31,  to  find  a?.  Ans.  a?  =  3. 

5.  Given  4a?+20— 6  =  34,  to  find  a?.  Ans.  a?  =  5. 

6.  Given  3a? -f  12-f  7a?  =  102,  to  find  a?.  Ans.  a?  =  9. 

7.  Given  10a?— 6a? -f  14  =  62,  to  find  a?.  Ans.  a?  =  12. 

8.  If  7a?—  14+5a?+20  =  246,  then  a?  =20.  For,  &c. 

9.  If  8a?+17— 5a?+3  =  100+10,  then  a?=30. 
10.  If  7a?—  14+3a?+35  =  450— 29,  then  a?  =  40. 

PROBLEMS. 

WtT  For  putting  questions  into  equations,  see  page  25. 
1.  What  number  is  that,  which,  with  5  added  to  it,  will 
be  equal  to  40  ? 

Stating  the  question,  a?  =  the  number 

a? +5,  =  after  adding 
Forming  the  equation,  *  a?+5  =  40 

Subtracting  5,  from  both,  a?  =  35. 


32  ALGEBRA.  ^SECTION  II. 

2.  A  man  being  asked  how  many  shillings  he  had,  answered, 
Add  15  to  their  number,  and  then  subtract  1,  and  the  remain- 
der will  be  64.     How  many  shillings  had  he  ? 

Stating  the  question,  #  =  number  of  shillings. 

a?+ 1*5  =  after  adding. 
#+15 — 1  =  after  subtracting. 
64  =  remainder. 
Forming  the  equation,  #+ 15  —  1  =  64 
Uniting  terms,  #+14  =  64- 

Subtracting  14  from  both  sides,  x  =  50. 

3.  What  number  is  that,  which  with  9  added  to  it,  will 
equal  23  ?     Ans.   14. 

4.  Divide  17  dollars  between  two  persons,  so  that  one  may 
have  4  dollars  more  than  the  other. 

Stating  the  question,  x  =  the  less  share. 

#+  4  =  the  greater. 
#+#-f-4  =  both  shares. 
17  =  both  shares. 
Forming  the  equation,  #  +  #+4=17 

Uniting  terms,  2#+4  =  17 

Subtracting  4  from  both  sides,       2#=  13 
Dividing  by  2,  #  =  6<£  share  of  one. 

17— 6£  =  10|  ••  the  other. 

5.  The  sum  of  the  ages  of  a  certain  man  and  his  wife  is  55 
years ;  and  his  age  exceeds  hers  by  7  years.  What  is  the 
age  of  each  ?     fS9^  Let  #  =  age  of  the  wife. 

Ans.  24  the  wife's;  31  the  man's. 

6.  A  is  5  years  older  than  B,  and  B  is  4  years  older  than 
C  ;  and  the  sum  of  their  ages  is  73  years.  What  is  the  age 
of  each  ? 

Stating  the  question,  #  =  C's  age. 

#+4  =  B's  age. 
#+4-f  5  =  A's  age. 
#+#  +  4+#  +  4  +  5  =  sum  of  all  of  them. 
Forming  the  equation,  #  +  #  +  4  +  #+4  +  5  =  73. 

Ans.  C  20  years,  B  24,  A  29. 


§  51.]  TRANSPOSITION   BY   SUBTRACTION.  33 

7.  Two  persons  were  candidates  for  a  certain  office,  where 
there  were  329  voters.  The  successful  candidate  gained  his 
election  by  a  majority  of  53.     How  many  voted  for  each  1 

Ans.   191  for  one,  and  138  for  the  other. 

8.  A,  B,  and  C,  would  divide  $200  among  themselves,  so 
that  B  may  have  $6  more  than  A ;  and  C  $8  more  than  B. 
How  much  must  each  have  ? 

Ans.  A  must  have  $60,  B  $66,  and  C  $74. 

9.  Divide  $1000  between  A,  B,  and  C  ;  so  that  A  shall 
have  $72  more  than  B,  and  C  $100  more  than  A. 

Ans.  Give  B  $252,  A  $324,  and  C  $424. 

10.  At  a  certain  election  1296  persons  voted,  and  the  suc- 
cessful candidate  had  a  majority  of  120.  How  many  voted 
for  each  ?  Ans.  588  for  one,  and  708  for  the  other. 

11.  A  father,  who  has  three  sons,  leaves  them  $8000,  spe- 
cifying in  his  will  that  the  second  shall  have  $500  more  than 
the  youngest,  and  that  the  eldest  shall  have  $1000  more  than 
the  second.  What  is  the  share  of  each  ?  Ans.  The  eldest 
had  $3500,  the  second  $2500,  the  youngest  $2000. 

12.  A  bin,  which  held  74  bushels,  was  filled  with  a  mixture 
of  corn,  rye,  and  oats.  In  it  there  were  15  bushels  of  rye 
more  than  of  corn ;  and  as  much  oats  as  both  corn  and  rye. 
What  was  the  quantity  of  each  ? 

Ans.   1 1  bushels  of  corn,  26  of  rye,  and  37  of  oats. 

13.  A  draper  bought  three  pieces  of  cloth,  which  together 
measured  159  yirds.  The  second  piece  was  15  yards  longer 
than  the  first,  and  ihe  third  24  yards  longer  than  the  second. 
Whal  was  the  length  of  each?  Ans.  The  first,  35  yards; 
the  second,  50  yards ;  the  third,  74  yards. 

C 


34  '  ALGEBRA.  [SECTION  III 


SECTION   III. 

TRANSPOSITION    BY    ADDITION. 

§  52.  It  is  frequently  found  that  some  quantity  has  been 
subtracted  from  the  a?'s  ;  as  in  the  equation    5a? — 44  =  76. 

§  53.  Now  if  we  had  a  dollar  minus  5  cents,  and  wished 
to  make  the  sum  just  a  dollar,  we  should  add  the  five  cents 
that  are  lacking.  So  in  the  above  equation,  we  have  5a? — 44; 
and  as  we  wish  to  make  it  just  5a?,  we  must  add  the  44  to  it; 
and  5a? — 44-|-44  is  the  same  as  5a?.  But  if  we  add  44  to 
one  member  of  the  equation,  we  must  also  add  as  much  to 
the  other  member  of  the  equation. 

So  that  adding  44  to  both  sides,  5a?— 44  +  44  =  76+44 
Or,  5a?  =  76+44 

Uniting  terms,  5a?  =  120 

Dividing  by  5,  a?  =   24. 

EQUATIONS.— SECTION  3. 

1.  Given  a?+ 14  +  3a?— 27  =  51,  to  find  x.  Ans.  a?  =16. 

2.  Given  3a?— 30— 2a?  =  46  —  7,  to  find  x.  Ans.  a?  =  69. 

3.  Given  9x—  41  +  6  =  88  —  6,  to  find  x.  Ans.  x—  13. 

4.  Given  20  +  3a?—  46  =  35— 4,  to  find  x.  Ans.  a?  =19. 

5.  Given  4a:      39  —  2a?  =  47,  to  find  x.  Ans.  a:  =  43. 

6.  Given  7a?+27—  46  =  65,  to  find  a?.  Ans.  a?  =  12. 

7.  Given  14a?— 55  —  8a?+ 14  =  85,  to  find  x.  Ans.  a?  =  21. 

PROBLEMS. 

19*  For  putting  questions  into  equations,  see  page  25. 
1.   What  number  is  that,  from  which  8  being  subtracted, 
the  remainder  is  45  ? 

Stating  the  question,  a?  =  the  number. 

a? — 8  =  when  8  is  subtracted 
Forming  the  equation,       a? — 8  =  45 
Adding  8  to  both  sides,  x  =  53 


$  53."]  TRANSPOSITION    BY    ADDITION.  35 

2.  What  number  is  that,  from  which  27  being  subtracted, 
the  remainder  is  41  ?  Ans.  68. 

3.  A  person  bought  two  geese  for  $1.40;  and  gave  16  cents 
more  for  one  than  he  did  for  the  other.  What  did  each  cost 
him  ? 

B^~  In  this  and  the  four  following  questions,  x  must  stand 
for  the  first  mentioned  quantity. 

Stating  the  question,  x  =  the  dearest. 

a?— 16  =  the  cheapest. 
x+x— 16  =  cost  of  both. 
140  =  cost  of  both. 
Forming  the  equation,         x-\- x— 16=  140 
Uniting  terms,  2x— 16  =  140 

Adding  16  to  both  sides,  2x=  156 

Dividing  by  2,  x  =   78. 

Ans.  78  cents,  and  62  cents. 

4.  Three  men,  who  are  engaged  in  trade,  put  in  $2600  as 
follows :  A  put  in  a  certain  sum ;  B,  $60  less  than  A ;  and 
C,  as  much  as  A  and  B,  lacking  $100.  What  was  each 
man's  share  ?  Ans.  A's  $705  ;  B's  $645  ;  C's  $1250. 

5.  A  purse  of  $8000  is  to  be  divided  among  A,  B,  and  C  ; 
bo  that  B  may  receive  $276  less  than  A,  and  C  $1112  less 
than  A  and  B  together.     What  is  each  man's  share  ? 

Ans.  A's  $2416;  B's  $2140;  C's  $3444. 

6.  A  gentleman  gave  to  two  beggars  49  cents  ;  giving  to 
the  first  15  cents  less  than  to  the  second.  How  many  cents 
did  each  receive?  Ans.  32  and  17. 

7.  A  man  leaves  $16000  to  be  divided  to  his  widow,  son, 
and  daughter,  in  such  a  manner  that  the  son  is  to  have  $2000 
less  than  the  widow,  and  the  daughter  $1000  less  than  the 
son.     What  is  the  share  of  each  ? 

Ans.   Widow,  $7000;  son,  $5000;  daughter,  $4000. 

8.  Divide  the  number  60  into  three  such  parts,  that  the 
first  may  exceed  the  second  by  8,  and  the  third  by  16. 

Ans.  28  ;  20  ;  and  12 


36  ALGEBRA.  [SECTION  III. 

9.  Three  men  having  found  a  purse  of  $160,  quarreled 
about  the  distribution  of  it.  After  the  quarrel,  it  was  found 
that  A  had  got  a  certain  sura,  and  that  B  had  $30  more  than 
A,  but  C  $50  less  than  A.     How  much  did  each  obtain  ? 

Ans.  A  $60;  B  $90 ;  C  $10. 

10.  Three  men,  A,  B,  and  C,  trade  in  company,  with  a 
stock  of  $3130;  of  which  B  puts  in  $350  more  than  A,  and 
C  $220  less  than  A.     What  was  the  capital  of  each  ? 

Ans.  A's$1000;  B's$1350;  C's  $780. 

11.  How  can  an  estate  of  $9931  be  divided  between  a 
widow,  son,  and  daughter,  in  such  a  manner  that  the  son 
shall  have  $592  less  than  the  widow,  and  $522  more  than 
the  daughter  ? 

WtBT"  After  knowing  the  son's  share,  how  can  the  daugh- 
ter's be  found  ? 

Ans.   Widow,  $3879  ;  son,  $3287  ;  daughter,  $2765. 

12.  A  father  divided  $12000  among  his  three  sons,  giving 
to  the  second  $1500  less  than  to  the  eldest,  and  $750  more 
than  to  the  youngest.     What  was  the  share  of  each  I 

Ans.  $5250  ;  $3750  ;  and  $3000. 

13.  A  father  has  willed  to  his  four  sons  $25200,  as  follows: 
To  D  a  certain  sum  ;  to  C  as  much  as  to  D,  lacking  $550;  to 
B  as  much  as  to  C,  together  with  $1550;  and  to  A  twice  as 
much  as  to  B,  lacking  $10000.     How  much  does    each  of 

them  receive  ? 

Ans.  A  $5100;  B  $7550 ;  C$6000;  D  $6550 


§  57.]       TRANSPOSITION  OF  THE  UNKNOWN  QUANTITY.  37 

SECTION  IV. 

TRANSPOSITION    OF    THE    UNKNOWN    QUANTITY. 

§  54.  We  have  found  that  when  any  term  has  the  sign  -}- 
it  may  be  removed  from  one  member  of  the  equation  to  the 
other,  if  we  take  care  to  change  the  sign  to  —  ;  for  this  has 
been  done  every  time  we  have  subtracted  a  term  from  both 
sides.     Thus,  in  the  equation 

#4-5  =  20; 
if  we  subtract  5  from   both    sides,   it  is  plain  that  the  first 
member  becomes  #,  and  the  last  member  becomes  20 — 5 ;  so 
that  the  equation  would  become 

#  =  20  —  5. 
§  55.  So  also  any  term  that  has  the  sign  —  may  be  re- 
moved from  one  member  to  the  other,  if  we  take  care  to 
change  the  sign  to  -}-.     Because  this  is  the  same  as  adding 
that  term  to  both  sides.     Thus,  in  the  equation 

#—5  =  20, 
if  we  add  5  to  both  sides,  the  first  member  becomes  x,  and 
the  last  member  becomes  20  +  5.     So  that  the  equation  be- 
comes 

a?  =  20  +  5. 

§  56.  When  we  remove  a  term  from  one  member  of  an 
equation  to  the  other  member,  we  say  that  we  transpose  that 
term  ;  and  the  operation  of  doing  it  is  called  transposition. 

§  57.  Any  term  may  be  transposed  from  one  member  of 
an  equation  to  the  other,  care  being  taken  to  change  the  sign 
when  we  change  the  side. 

§  58.  It  was  stated   in  §  25,  that   an  equation  must  be 


Questions.  How  can  a  positive  quantity  be  removed  from  one 
member  of  an  equation  to  the  other]  Why?  How  may  a  negative 
quantity  be  removed  from  one  side  to  the  other  ?  Why  ?  What  do 
we  call  this  method  of  removing  a  term  i  What  care  is  required  in 
transposing?     Why  are  we  ever  obliged  to  transpose  1 

4 


38  ALGEBRA.  [SECTION  IV 

brought  so  that  the  unknown  quantity  will  occ  ipy  one  mem 
ber  of  the  equation,  and  the  known  quantities  embrace  the 
other  member.  And,  as  it  frequently  happens  that  the  un- 
known quantities  are  on  both  sides,  we  are  obliged  to  resort  to 
transposition  in  order  to  make  one  side  -free  from  them.  And 
likewise,  it  is  often  necessary  to  transpose  known  quantities 
from  the  member  which  contains  the  unknown  quantity. 

RULE    FOR    TRANSPOSING. 

§  59.  In  transposing,  it  is  generally  best  to  write  first  the 
unknown  quantify  that  is  already  on  the  left ;  and  then  bring 
over  all  those  which  are  on  the  right,  if  there  are  any  there. 
And  in  transposing  the  known  quantifies  to  the  right  hand 
member,  write  those  that  are  already  there,  and  then  trans- 
pose after  them  what  known  quantities  there  are  in  the  left. 

EQUATIONS.— SECTION  4. 

1.  Reduce  the  equation      \x —  14  =  3#-f-12. 
Transposing  3*,  >  ^_^  _  ^ 
Transposing  14,3 

Uniting  terms,  x  =  26. 

2.  Given  21 — 7x  =  40 — 11#,  to  find  x.    Ans.  a?  =  4f. 

3.  Given  40  —  6z  ==  136—  14*,  to  find  z.    Ans.  z  =  12. 

4.  Given  Sy —  4  =  y+ 12,  to  find  y.  Ans.  y  =  8. 

5.  Given  bx — 15  mi  2#+6,  to  find  x.         Ans.  x  =  7. 

6.  Given  40  —  6x  — 16  =•- 120  —  14r,  to  find  x.  Ans.  12. 

7.  Given  4  —  9y  =  14  —  lly,  to  find  y.      Ans.  y  =  5. 

8.  Given  x+ 18  =  Sx  —  5,  to  find  x. 
Transposing,  x  —  3a?  =  —  5  — 18 
Uniting  terms,  — 2x  =  — 23 
Dividing  by  2,  —  a?=  — 11£ 


Question.  In  what  order  do  we  write  the  terms  when  we  are 
••ansposing  ? 

*  The  pupil  will  recite  the  left  hand  member  of  this  line,  for  trans- 
posing 3x. 


§  63.]       TRANSPOSITION  OF  THE  UNKNOWN  QUANTITY.  39 

§  60.  It  is  of  no  consequence  what  sign  accompanies  the 
final  result ;  as  the  magnitude  of  the  quantity  is  not  affected 
by  the  sign.  If  we  remember  that  +  is  understood  and  may 
be  written  with  every  positive  quantity,  it  will  be  very  evident 
that  the  equation  — x  =  — 1 1£  is  just  as  good  as  the  equation 
+  x  =  +11 5.  In  both  cases,  the  quantity  x  is  equal  to  the 
number  11$. 

§  61.  In  the  result  of  the  last  question,  11|  may  be  trans- 
posed to  the  first  member ;  and  x  may  be  transposed  to  the 
last  member.  Of  course,  this  will  change  the  signs ;  and  the 
equation  will  become  1  \\  =  x.  And  if  1 15  =  a?,  it  is  evident 
that  x  =  1 1£.     This  coincides  with  what  was  shown  in  §  60. 

§  62.  From  what  has  just  been  said,  we  see  that  all  the 
terms  of  each  member  may  be  transposed,  so  that  the  sign  of 
each  term  may  be  changed ;  and  still  the  equation  shall  retain 
the  same  members  as  at  nrst,  though  differently  placed. 
Hence,  it  is  immaterial  which  member  is  written  first.  And 
also,  in  any  equation  the  signs  of  all  the  terms  may  be 
changed  without  affecting  the  equality. 

§  63.  It  is  evident  that  all  the  terms  of  one  member  may 
be  transposed  to  the  other  member.  When  this  has  been 
done,  the  member  fr?m  which  the  terms  have  been  trans- 
posed becomes,  0.  Thus  the  equation  x=ll£,  may  be 
made  x — ll£=0;  where — ll£  balances  x. 

PROBLEMS. 

WSr  After  the  equation  has  been  formed  by  §  47,  it 
must  be  transposed  by  §  59. 

1.  A  man  has  six  sons,  whose  successive  ages  differ  by 
4  years ;  and  the  eldest  is  three  times  as  old  as  the  young- 
est.    What  are  their  ages  ? 


Questions.  What  sign  must  accompany  the  answer?  Explain. 
Explain  by  transposition.  To  what  extent  may  the  signs  be  chvinged1 
Why?     To  what  extent  may  the  terms  be  transposed  1     Why? 


40 


ALGEBRA. 


[SECTION  IV 


Stating  the  question, 


x  =  age  of  the  youngest 


#  +  4  =  "  "  next 

#  +  4  +  4  ==  "  "  next 

#  +  4  +  4  +  4  =  "  "  next. 

#+4  +  4-1-4  +  4-=  "  "  next. 

#+4  +  4  +  4  +  4  +  4  =  "  "  eldest. 

3x  =  "  "  eldest. 
Forming  the  equation,  #+4+4+4+4+4  =  3a? 


Uniting  terms, 
Transposing  the  3#  and  20, 
Uniting  terms, 
Dividing  by  2, 
or 


#+20  =  3# 
#  — 3#  =  —20 
—  2*  =  —  20 
—  #  =  — 10 

x  =  1 0  age  of  the 

[youngest. 

2.  A  person  bought  two  horses,  and  also  a  hundred  dollar 
harness.  The  first  horse,  witl*  the  harness,  was  of  equal 
value  with  the  second  horse.  But  the  second  horse  with  the 
harness  cost  twice  as  much  as  the  first.  What  was  the  price 
of  each  horse  ? 

Stating  the  question,  x  =  price  of  the  first. 

#+ 100  =  price  of  the  second. 
#+100+100  =  2d  horse  harnessed. 
2x  =  twice  price  of  first. 
Forming  the  equation,     x  + 1 00  + 1 00  =  2x 
Transpos.  from  both  members,  x — 2#  =  — 100 — 100 
Uniting  terms,  — x  =  — 200 

Or  x  =  200  &c. 

3.  A  privateer,  running  at  the  rate  of  10  miles  an  hour, 
discovers  a  ship  18  miles  off,  sailing  at  the  rate  of  8  miles  an 
hour.  How  many  hours  can  the  ship  run  before  she  will  be 
overtaken  by  the  privateer? 

H©^  In  x  hours,  the  privateer  will  go  10a?  miles,  which 
is  the  whole  distance.  In  the  same  time,  the  ship  will  go 
Sx  miles.  But  the  ship  had  already  gone  18  miles,  which, 
added  to  the  Sx,  will  make  the  whole  distance. 


5  63.]  TRANSPOSITION.  41 

4.  A  gentleman  distributing  money  among  some  poor  peo- 
ple, found  that  if  he  undertook  to  give  5s.  to  each,  he  would 
lack  10s.  Therefore,  he  gave  only  4s.  to  each,  and  finds  that 
he  has  5s.  left.     How  many  persons  were  there  ? 

W^~  It  will  be  found  that  5a? — 10=  his  money  by  the 
first  supposition,  and  4x+5  =  the  money  by  the  last  suppo- 
sition. Ans.  15. 

5.  I  once  had  $84  in  my  possession ;  and  I  gave  away  so 
much  of  it,  that  what  I  have  now  equals  three  times  as  much 
as  I  gave  away.     How  much  did  I  give  away  ? 

ffl&T  If  I  gave  away  $#,  then  $84  —  x  will  be  what  re- 
mains. Ans.  $21. 

6.  A  certain  sum  of  money  was  shared  among  five  persons, 
A,  B,  C,  D,  and  E.  Now  B  received  $10  less  than  A  ;  C, 
$16  more  than  B ;  D,  $5  less  than  C  ;  E,  $15  more  than  D. 
And  it  was  found  that  the  sum  of  the  shares  of  the  first  three 
put  together,  were  equal  to  the  sum  of  the  shares  of  the  other 
two.     How  much  did  each  man  receive  ? 

Ans.  A  $21  ;  B  $11  ;  C  $27  ;  D  $22 ;  E  $37. 

7.  A  person  wishes  to  give  3  cents  apiece  to  some  beggars, 
but  finds  that  he  has  not  money  enough  by  8  cents.  He 
gives  them  2  cents  apiece  and  has  3  cents  left.  How  many 
beggars  were  there  ?  Ans.   11. 

8.  A  courier  who  had  started  from  a  certain  place  10  hours 
ago,  is  pursued  by  another  from  the  same  place,  and  on  the 
same  road.    The  first  goes  4  miles  an  hour,  and  the  second  9 
In  how  many  hours  will  the  pursuer  overtake  the  first? 

fgy  If  the  pursuer  goes  x  hours,  the  first  must  go 
x-f  10  hours.  But,  as  both  go  from  the  same  place,  tha 
distance  that  each  goes  must  be  the  same. 

In  generalization  see  page  113. 


42  ALGEBRA.  [SECTION  V. 


SECTION  V. 

MULTIPLICATION  OF  COMPOUND   QUANTITIES 

BY  SIMPLE  QUANTITIES. 

v 

§  64.  Suppose  you  purchase  8  melons  at  7  cents  apiece, 
and  afterwards  find  that  you  must  give  5  cents  apiece  more 
for  them.  In  this  case  you  pay  8  times  7  cents,  and  also  8 
times  5  cents ;  that  is,  first,  56  cents,  and  afterwards  40  cents. 

§  65.  Let  us  apply  this  principle  to  Algebra.  You  pay  in 
all,  8  times  (7  +  5,)  which  =56+10.  Which  shows  that  in 
multiplying  a  compound  quantity,  you  multiply  each  term 
by  the  multiplier. 

We  can  easily  see  that  this  operation  will  give  the  right 
answer;  for  in  the  case  of  the  melons,  they  cost  12  cents 
apiece,  and  therefore  their  whole  cost  was  8  times  12  cents 
which  =96  cents.  But  the  answer  just  obtained,  56+40 
=  96. 

§  66.  But  suppose  that  after  you  had  paid  7  cents  apiece, 
a  deduction  of  5  cents  apiece  was  made.  The  whole  cost 
would  then  be  8  times  (7  —  5,)  which  =56  —  40.  And  this 
agrees  with  the  truth ;  for  you  first  paid  56  cents,  and  after- 
wards 40  cents  were  deducted. 

W£T  — 5  multiplied  by  8,  signifies  that  — 5  is  to  be  added 
8  times.     Therefore  retaining  the  sign,  (§29),  we  add  — 40. 

§  67.  This  shows  that  +  multiplied  by  +,  produces  +  ; 
and  —  multiplied  by  +,  produces  — . 

EXAMPLES. 

1.  Multiply  a; +4,  by  3. 

Operation.         a? +4 


Ans.  'Sx  +12. 


Questions.  How  do  we  multiply  a  compound  quantity]  Explain 
why.  How  do  the  signs  of  the  answer  correspond  with  the  quantity 
that  is  multiplied  1     Explain  for  the  — , 


5  69.1         MULTIPLICATION  OF  COMPOUND  QUANTITIES.  43 

2.  Multiply  12  +  a?,  by  5.  Ans.  60  +  5a?. 

3.  Multiply  x— 10,  by  8.  Ans.  8a?  — 80. 

4.  Multiply  126— a-,  by  4.  Ans.  504— 4a:. 

5.  Multiply  a? +  8,  by  6. 

6.  Multiply  40  +  a*,  by  10.  , 

7.  Multiply  a?— 32,  by  9. 

8.  Multiply  52— x,  by  12. 

9.  Multiply  2a? +  14,  by  7. 

10.  Multiply  27  + 3a?,  by  14. 

11.  Multiply  3a?— 62,  by  15. 

12.  Multiply  97  — 4a?,  by  12. 

13.  Multiply  a?+7— y,  by  7. 

14.  Multiply  Sx+y— 12,  by  8. 

15.  Multiply  2a?— 3y— 6,  by  6. 

16.  Multiply  3a?— 12+y,  by  5. 

§  68.  Franciscus  Vieta,  a  Frenchman,  introduced  about  the 
year  1600,  the  vinculum  or  a  straight  line  drawn  over  the  top 
of  two  or  more  quantities  when  it  is  wished  to  connect  them 
together.  Thus,  a  +  4x3,  signifies  that  both  a?  and  4  are  to 
be  multiplied  by  3. 

§  69.  In  1629,  Albert  Girard,  a  Dutchman,  introduced  the 
parenthesis  as  a  convenient  substitute,  in  many  cases,  for  the 
vinculum.  Thus,  (a?+4)x3,  is  the  same  as  a*  +  4x3  ;  and  is 
read,  a? +4,  both  x3.  If  there  are  more  than  two  terms  under 
the  vinculum,  we  say,  after  repeating  those  terms,  all,  &c. 
Thus,  (a?+y)  x(« — b+c),  is  read  x-\-y  both  into  a—b  +  c 
all.     See  also  §  103. 

Questions.  What  is  a  vinculum  1  How  is  the  parenthesis  used  1 
How  is  a  compound  quantity  read  when  embraced  by  a  vinculum  or 
parenthesis  1. 


44  ALGEBRA.  lSECTI0N  T» 

EQUATIONS.  —  SECTION  5. 

1.  Given  x  — 9x11  =  121,  to  find  a?. 

Writing  the  equation,  x— 9  X 1 1  =  121 

Performing  the  multiplication,    \\x — 99  =121 
Transposing  and  uniting,  llx  =  220 

Dividing  by  11,  x  =  20. 

2.  Given  (#+7) x6  =  54,  to  find  a?.  Ans.  x  =  2. 


3.  Given  12  +  xx5  =  100,  to  find  x.  Ans.  8. 

4.  Given  x— 9  x8  =  96,  to  find  x.  Ans.  21, 


5.  Given  367— 3#x5  =  920,  to  find  x.  Ans.  61. 

6.  Given(8+a:)x2+14  =  72,  to  find  a?. 

10°  The  pupil  must  understand  that  14  is  not  a  part  of 
the  multiplier,  because  there  is  the  sign  -f  between  it  and  the 
ultiplier.  Ans.  21. 

7.  Given  (15+*) x3— 27  =  48,  to  find  x.       Ans.  10. 

8.  (112— 2*)x3  =  (2x— 7)x4,  to  find  x.        Ans.  26. 

9.  (3z+14)x4  =  (78— a*)x5,  to  find  ic.  Ans.  19}|. 


10.  2x  +  8x5  =  (32 +.r)x3,  to  find  a\  Ans.  8. 

11.  (3x— 14)x7  =  (17— x)x6,  to  find  x.  Ans.  7Jf 


12.  120— 3xx2  =  (4x— 6)x9,  to  find  a?.  Ans.  7. 

PROBLEMS. 

1.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money  in 
trade  ;  A  gains  $126,  and  B  loses  $87  ;  and  now  A's  money 
is  double  of  B's.     What  did  each  lay  out? 

Stating  the  question,  x,  =  the  sum  for  each. 

a?+126  =  A's  sum  now. 
x — 87  =  B's  sum  now. 
2x— 174  =  the  double  of  B's 
Forming  the  equation,  #-r-126  =  2:r— 174. 

Transposing  and  uniting,  — a?=  — 300. 

Changing  signs,  x  =  300  the  answei 


§  69.  j        MULTIPLICATION  OF  COMPOUND  QUANTITIES.  45 

2.  A  person,  at  the  time  he  was  married,  was  3  times  as 
old  as  his  wife;  but  after  they  had  lived  together  15  years, 
he  was  only  twice  as  old.  What  were  their  ages  on  their 
wedding  day? 

Stating  the  question,  x  =  the  wife's  age. 

Sx  =  the  man's  age* 
a?+  15=  the  wife's  after  15  years 
3a?+15  =  the  man's  after  15  years. 
2a?+30  =  twice  the  wife's  age. 
Forming  the  equation,      3#+ 1 5  =  2.T+30 
Transposing  and  uniting,  x=  15  the  wife's  age. 

Sx  =  45  the  man's  age. 

3.  A  man  having  some  calves  and  some  sheep,  and  being 
asked  how  many  he  had  of  each  sort;  answered  that  he  had 
twenty  more  sheep  than  calves,  and  that  seven  times  the  num- 
ber of  calves  was  equal  to  three  times  the  number  of  sheep. 
How  many  were  there  of  each  ? 

Ir^T3*  If  x=  number  of  calves,  then  x  +  20=  number  of 
sheep.  Ans.    15  calves,  and  35  sheep. 

4.  Two  persons,  A  and  B,  having  received  equal  sums  of 
money,  A  paid  away  $25,  and  B  paid  away  $60 ;  and  then 
it  appeared  that  A  had  just  twice  as  much  money  left  as  B. 
What  was  the  sum  that  each  received  ?  Ans.  $95. 

5.  Divide  the  number  75  into  two  such  parts,  so  that 
three  times  the  greater  may  exceed  7  times  the  less  by  15. 

I'lT*  If  x=  the  greater  then  75— x  =  the  less;  and  3a? 
will  =  (7  times  the  less)  +  15.  Ans.  54  and  21. 

6.  The  garrison  of  a  certain  town  consists  of  125  men, 
partly  cavalry  and  partly  infantry.  The  monthly  pay  of  a 
horse  soldier  is  $20,  and  that  of  a  foot  soldier  is  $15;  and 
the  whole  garrison  receives  $2050  a  month.  What  is  the 
number  of  cavalry,  and  what  of  infantry  ? 

If  j  =  number  of  cavalry,  then  20#  =  the  whole  pay 
of  cavalry,  &c.  Are.  35  cavalry,  and  90  infantry 


46  ALGEBRA.  [SECTION  V. 

7.  A  grocer  sold  his  brandy  for  25  cents  a  gallon  more  than 
he  asked  for  his  wine ;  and  37  gallons  of  his  wine  came  to 
as  much  as  32  gallons  of  his  brandy.  What  was  each  per 
gallon?  Ans.  $1.60  for  wine;  and  $1.85  for  brandy. 

8.  A  wine  merchant  has  two  kinds  of  wine ;  the  one  costs 
9  shillings  a  gallon,  the  other  5.  He  wishes  to  mix  both 
wines  together,  so  that  he  may  have  50  gallons  that  may  be 
sold  without  profit  or  loss  for  8  shillings  a  gallon.  How  much 
must  he  take  of  each  sort  ? 

SZW  There  are  50  gallons  of  both  kinds,  and  after  rina- 
ing  the  cost  by  the  kinds,  the  whole  mixture  will  be  worth 
50  times  8  shillings. 

Ans.  37  d  gallons  of  the  best;  and  125  of  the  poorer. 

9.  A  gentleman  is  now  40  years  old,  and  his  son  is  9  years 
old.  In  how  many  years,  if  they  both  live,  will  the  father  be 
only  twice  as  old  as  his  son  ? 

JO1*  In  x  years  he  will  be  40-)-#>  and  his  son  9-\-x. 

Ans.  22  years. 

10.  A  man  bought  20  oranges  and  25  lemons  for  $1.95. 
For  each  of  the  oranges  he  gave  3  cents  more  than  for  a 
lemon.     What  did  he  give  apiece  for  each  ? 

Ans.  3  cents  for  lemons  ;  6  cents  for  oranges. 

11.  A  man  sold  45  barrels  of  flour  for  $279  ;  some  at  $5  a 
barrel,  and  some  at  $8. .  How  many  barrels  were  there  of 
each  sort?  Ans.  27  at  $5  ;  and  18  at  $8. 

12.  Says  John  to  William,  I  have  three  times  as  many 
marbles  as  you.  Yes,  says  William;  but  if  you  will  give  me 
20,  I  shall  have  7  times  as  many  as  you.  How  many  has 
each? 

!Cr"  Let  x  =  William's,  and  3a?  =  John's.  Then  after 
the  change,  #+20=  William's,  and  3x  —  20  =  John's. 

Ans.  John  24  ;   William  8. 

13.  A  person  bought  a  chaise,  horse,  and  harness,  for  $440. 
The  horse  cost  him  the  price  of  the  harness,  and  $80  more ; 
and  the  chaise  cost  twice  the  price  of  the  horse.  What  did 
he  give  for  each  ? 

Ans.  For  the  harness  $50  ;  horse  $130 ;  chaise  $260 


§  69.]         MULTIPLICATION  OF  COMPOUND  QUANTITIES.  47 

14.  Two  men  talking  of  their  ages,  the  first  says,  Your  age 
is  18  years  more  than  mine,  and  twice  your  age  is  equal  to 
three  times  mine.     What  is  the  age  of  each  ? 

Ans.  Youngest,  36  years  ;  eldest,  54  years. 

15.  A  boy  had  41  apples  which  he  wished  to  divide  among 
three  companions  as  follows :  to  the  second,  twice  as  many 
as  to  the  first,  and  3  apples  more ;  and  to  the  third,  three 
times  as  many  as  to  the  second,  and  2  apples  more.  How 
many  did  he  give  to  each  ? 

.  Ans.  To  the  first,  3  ;  second,  9;  third,  29^, 

^l  16.  How  many  gallons  of  wine,  at  9  shillings  a  gallon, 
must  be  mixed  with  20  gallons  at  13  shillings,  so  that  the 
mixture  may  be  worth  10  shillings  a  gallon  ?    Ans.  60  gallons. 

17.  Two  persons,  A  and  B,  have  each  an  annual  income 
of  $400.  A  spends,  every  year,  $40  more  than  B ;  and,  at 
the  end  of  4  years,  they  both  together  save  a  sum  equal  to  the 
yearly  income  of  either.     What  do  they  spend  annually  ? 

Ans.  A,  $370 ;  B,  $330. 

18.  A  farmer  wishes  to  mix  rye  worth  72  cents  a  bushel, 
with  oats  worth  45  cents  a  bushel,  so  that  he  may  have  100 
bushels  worth  54  cents  a  bushel.  How  many  bushels  of  each 
sort  must  he  take  ?  Ans.  33 £  of  rye,  and  66|  of  oats. 

In  generalization  see  page  136. 


48  ALGEBRA.  [SECTION  TI. 

SECTION   VI. 
FRACTIONS. 

§  70.  All  the  division  which  the  pupil  has  as  yet  performed, 
has  been  the  division  either  of  numeral  quantities,  or  of  the 
numeral  co-efficients.  But  in  Algebra,  it  is  frequently  neces- 
sary to  divide  literal  quantities.  For  example,  after  having 
made  x  to  stand  for  an  unknown  quantity,  we  may  wish  to 
find  the  half  of  x,  or  the  third  of  x,  or  the  fourth  of  x,  &c. 

§  71.  In  common  arithmetic,  if  we  wish  to  divide  1  by  2, 
we  do  it  by  writing  2  under  the  1  ;  thus,  5.  So  if  we  wish 
to  divide  2  by  3,  we  write  3  under  the  2;  thus,  I.  In  the 
same  manner,  2  divided  by  5  is  written  -f ;  3^-4  is  written  | ; 
6-i-7  is  written  ^.  The  quantities  that  are  obtained  by  di- 
viding in  t  is  manner,  are  called  J ract  ion*. 

§  72.  In  Algebra,  we  most  generally  make  use  of  this 
method  of  dividing ;  especially  when  we  divide  literal  quan- 
tities. Or,  in  other  words,  we  divide  a  literal  quantity  by 
writing  the  divisor  under  the  dividend,  with  a  straight  line 
between  them;  thus,  x  divided  by  2,  is  written  £;  and  is 
read  j-ha/f  x-r-S,  is  written  *  ;  and  is  read,  x-third ;  x-r-4, 
is  written  J,  and  is  read  x-fourth;  ^x—A  =  3~,  and  is  read, 
Sx-fourth. 

§  73.  The  two  separate  numbers  that  we  employ  in  writing 
a  fraction,  are  called  terms.  The  upper  term  is  called  the 
numerator,  and  the  lower  term  is  called  the  den  mi/iat  >r. 
Thus,  in  the  fraction  f ,  we  call  x  the  numerator,  and  3  the 
denominator. 

§  74.  If  the  one-third  of  x  is  f ,  it  is  evident  that  f  of  x, 
is  two  times  as  much;  that  is  I*.     If  J  of  a?  is  |,  then  |  of 


Questions.  Can  literal  quantities  be  divided  1  In  what  manner  1 
Which  part  of  a  fraction  is  the  denominator  1  Which  is  the  nume- 
rator1    What  are  they  both  called? 


§  77.]  FRACTIONS.  49 

x  is  3|.  Whence  the  rule  to  multiply  a  whole  number  by  a 
fraction,  is,  to  multiply  the  whole  number  by  the  numera- 
tor,  and  divide  by  the  denominator;  as  £  of  x  is  4£ ;  f  of  y 
is?*;  |  of  a  is  2f 

Examples.  The  pupil  may  multiply  a,  a;,  and  y,  each  of 
them  by  f ;  and  then  by  £ ;  and  then  by  f ,  |,  £ ,  f ,  $,.|,  suc- 
cessively. 

§  75.  As  we  can  multiply  a  number  of  parts  as  well  as  a 
number  of  wholes,  and  as  the  denominator  is  nothing  more 
than  the  name  of  the  parts ;  it  is  plain  that  to  multiply  a  frac- 
tion, we  multiply  the  numerator,  and  retain  the  denomina- 
tor without  alteration.  Thus,  2  times  |  is  f ;  3  times  |-  is 
£ ;  2  times  f  is  f ;  4  times  §  is  f ;  &c. 

Examples.  Multiply  each  of  the  following  fractions  by  2, 
then  by  3,  and  then  by  4.     f,  |,  f ,  4,  f ,  f ,  f ,  £,  f ,  §,  |,  * 

3x     4x 

T'  1' 

§  76.  As  we  know  that  2  halves  =  a  whole,  we  readily 
conclude  that  4  halves  =  2  wholes ;  and  that  6  halves  =  3 
wholes,  &c.  Likewise,  because  3  thirds  =  a  whole,  6  thirds 
must  equal  2  wholes  ;  and  9  thirds  must  equal  3  wholes.  In 
the  same  manner  8  fourths  =  2  wholes ;  20  fifths  =  four 
wholes;  18  thirds  =  6  wholes,  &c.  Such  fractions  as  {, 
2£,  ^,  &c,  are  called  improper  fractions, 

§  77.  Hence,  in  order  to  find  how  many  whole  ones  there 
are  in  any  number  of  halves,  we  have  only  to  see  how  many 
times  two  halves  are  contained  in  that  number.  Thus,  in  10 
halves  there  are  as  many  whole  ones  as  there  are  2  halves 
contained  in  10  halves  ;  which  is  5.  In  the  same  manner,  in 
12  thirds  there  are  as  many  whole  ones  as  there  are  3  thirds 
contained  in  12  thirds  ;  which  is  4. 


Questions.     How  do  we  multiply  a  whole  number  by  a  fraction? 
Give  the  answers  to  the  examples.   How  do  we  multiply  a  fraction? 
Examples.     What  are  improper  fractions?     How  can  an  improper 
"raction  be  changed  to  a  whole  number? 
D  5 


50  ALGEBRA.  [SECTION  VI. 

§  78.  Thus  we  have  the  rule,  to  change  an  improper 
fraction  to  a  whole  number,  divide  the  numerator  by  the 
denominator. 

When  the  answer  consists  of  an  integer  and  a  fraction,  it  is 
called  a  mixed  number. 

Examples.     1.  How  many  whole  ones  in  -|? 

Ans.  8-~3  =  2$. 

2.  How  many  whole  ones  in  } ?   %1  ^\  J?   f? 

3.  How  many  whole  ones  in' »?   ??   g?   *1   {?? 

4.  How  many  whole  ar's  in  Jt   $5?   8p?   ^? 

5.  How  many  whole  x's  in  if?   ?|x?   Jp!   5|«? 

6.  How  many  whole  a^'s  in  3  times  *?? 

7.  How  many  whole  x's  in  4  times  \  ? 

8.  How  many  whole  a*'s  in  5  times  J ?v^ 

§  79.  If  we  have  the  quantity  f ,  we  know  that,  as  it  takes 
5  fifths  to  make  a  whole  one,  it  will  take  5  times  this  quan- 
tity to  make  a  whole  x.  Therefore,  if  we  multiply  *  by  5, 
we  shall  obtain  ^,  or  exactly  x.  If  we  multiply  ^  by  3,  we 
shall  obtain  ^,  or,  which  is  the  same,  x.  If  we  multiply  f  by 
4,  we  shall  obtain  4|,  or  x. 

§  80.  As  4  times  f  is  equal  to  x ;  then  4  times  2|  must  be 
equal  to  twice  as  much,  or  2x ;  and  4  times  j  must  be  three 
times  as  much,  or  3a?.  As  3  times  £  is  equal  to  x;  so  3  times 
5?  must  be  twice  as  much,  or  2x. 

§  81.  Any  fraction  when  multiplied  by  the  number  which 
is  the  same  as  the  denominator,  will  produce  a  quantity 
which  is  the  same  as  the  numerator.     Thus, 
5|X4  =  5.t;         ^xS=7x. 

We  shall  be  able  to  make  use  of  this  principle  in  the  solu- 
tion of  many  equations,  if  we  operate  in  accordance  with  the 
following  axiom  or  self-evident  truth. 

Questions.    What  is  a  mixed  number1?    What  is  obtained  by  mul 
tiplying  a  fraction  by  a  number  that  is  the  same  as  the  denominator  * 


§  86.]  FRACTIONS.  51 

§32.  Jf  equals  be  multiplied  by  the  same,  their  products 
will  be  equal.    Thus,  if  x  =  10,  then  2x  =  20 ;  4x  =  40,  &c. 

§  83.  It  is  evident  [§76]  that  each  of  the  following  frac- 
tions, §,  |,  £,  f ,  |,  },  f ,  |,  &c,  is  equal  to  1.  Therefore, 
they  must  be  equal  to  one  another.  Also,  each  of  the  follow- 
ing fractions,  f ,  J,  f ,  |,  ^2,  &c,  is  equal  to  2 ;  and  conse- 
quently they  are  all  equal  to  one  another.  In  the  same  man- 
ner, we  may  make  many  fractions  that  will  equal  3 ;  and  so 
of  any  other  number. 

Fractions  that  are  equal  to  one  another  but  have  different 
terms,  are  called  equivalent  fractions. 

§  84.  Let  us  take  from  the  first  set  of  the  above  fractions, 
§  and  *  which  are  equal  to  one  another.  We  see  that  both 
the  numerator  and  denominator  in  the  last  fraction  are  twice 
as  much  as  in  the  first.  We  find  the  same,  by  taking  from  the 
second  set,  the  equal  fractions  \ ,  | ;  and  also  -|  and  £ ;  and 
also  f  and  j2. 

Again,  in  the  equal  fractions,  §  and  f ,  we  find  each  term  in 
the  last  fraction  three  times  as  great  as  the  correspondent 
term  in  the  first  fraction.  The  same  may  be  observed  in  the 
fractions  |  and  -§■ ;  and  also  in  f  and  £  and  also  in  f  and  f . 

§  85.  By  pursuing  this  investigation,  we  shall  find  that 
whenever  we  multiply  both  the  numerator  and  the  denomi- 
nator by  the  same  number,  no  matter  what  that  number  may- 
be, the  fraction  made  by  that  multiplication  will  be  equal  in 
value  to  the  first  fraction.  Hence,  there  is  an  equality  between 
the  value  of  the  following  fractions,  ^,  f ,  £,  f ,  T5„,  T\,  <fcc. ; 
and  also  between  the  following,  {,  },  f,  T\,  -fa  T\,  &c. 

§  86.  The  principle  just  explained,  leads  to  another  which 
is  of  much  importance.  Suppose  we  multiply  by  8,  both 
terms  of  the  fraction  f  ;  and  obtain  \\.  It  is  plain  that  both 
terms  of  the  fraction  *£•  can  be  divided  by  8,  to  bring  the  frac- 

Queslions.  How  can  we  multiply  without  destroying  the  equa- 
tion 1  What  are  equivalent  fractions  ?  What  will  be  the  effect  of 
multiplying  both  terms  of  a  fraction  by  any  number? 


52  ALGEBRA.  [SECTION  VI. 

tion  back  to  I.  So  also,  if  both  terms  in  |  be  multiplied  by 
6,  the  fraction  will  be  -*£,  which  means  just  as  much  as  ?  ; 
and,  of  course,  if  both  terms  in  ±%  be  divided  by  6,  the  frac- 
tion will  be  brought  back  to  £,  which  is  equal  to  J-f  •  So,  in 
general,  if  we  divide  both  the  numerator  and  the  denomina- 
tor of  a  fraction  by  the  same  number,  we  have  a  new  frac- 
tion which  will  be  equal  to  the  first.  Thus,  T8^  may  be 
changed  to  \ ;   if  to  J  ;   ^  to  f 

§  87.  It  is  evident,  that  of  several  fractions  of  equal  value, 
that  which  has  the  least  denominator  is  the  most  easily  un- 
derstood. Thus,  |  of  an  apple  is  much  better  known  at  first 
sight,  than  ff  of  an  apple.  And  when  a  fraction  is  brought 
to  as  small  a  denominator  as  it  can  be  changed  to,  we  say 
it  is  reduced  to  its  lowest  terms. 

§88.  In  order  to  reduce  a  fraction  to  its  lowest  terms,  di- 
vide both  the  numerator  and  the  denominator  by  any  num- 
ber that  will  divide  each  without  a  remainder.  Thus,  in 
the  fraction  1^y,  both  terms  may  be  divided  by  5,  by  which 
we  obtain  £-f ;  and  both  terms  of  this  last  fraction  may  be  di- 
vided by  3,  by  which  we  obtain  £. 

EQUATIONS.  — SECTION  6. 

PROBLEMS. 

1.  In  an  orchard,  |  of  the  trees  bear  apples,  ^  of  them  bear 
pears,  -^  bear  plums,  and  81  bear  cherries.  How  many  trees 
are  there  in  the  orchard  ;  and  how  many  of  each  sort  ? 

Stating  the  question,  x  =  number  of  trees. 

f  =  apple  trees. 
£  =  pear  trees, 
•fy  =  plum  tree. 
81  =  cherry  trees. 
All  these  trees  together  =  the  whole  number 
Forming  the  equation,         x  =  £-r-f-f-f^  +  81. 

Questions.  Suppose  we  divide  both  by  any  number  ?  How  do 
we  reduce  a  fraction  to  its  lowest  terms? 


§  88.J  FRACTIONS.  53 

Now,  we  know  that  if  we  multiply  |  by  4  we  obtain  a? 
alone  ;  that  is,  we  destroy  the  fraction,  and  make  it  a  whole 
number.  And  we  know,  that  if  we  multiply  the  first  member 
by  4,  and  also  the  last  member  by  4,  we  shall  not  destroy  the 
equation.  See  §  82.  We  will  therefore  multiply  both  mem- 
bers by  4,  for  the  purpose  of  destroying  the  fraction  in  the 
first  term.     It  will  then  become 

.      4a?  =  a?+f+-f£+324. 
Next,  we  will  multiply  both  members  by  5,  to  destroy  the 
fraction  in  the  second  term.     This  will  make 
20a?  =  5a? + 4a? + ^f  + 1620. 
Then  we  will  multiply  by  11,  to  destroy  the  remaining 
'fraction,  which  will  make 

220a?  =  55a?  -j-  44a? + 40a?  +17820. 
Transposing  and  uniting,  81a?  =  17820. 

Dividing  by  8 1 ,  a?  =  220.     the  Ans. 

2.  In  a  certain  school,  £  of  the  boys  learn  mathematics,  | 
of  them  study  Latin  and  Greek,  and  6  study  English  gram- 
mar.    What  is  the  whole  number  of  scholars  ? 

ICT*   After  the  question  is  stated,  the  equation  will  be- 
come a:  =  £4-54- 6 
Multiplying  by  5,  to  de-     j             5a?  =  a? 44? 4- 30 

stroy  the  first  fraction,      5 
Multiplying  by  4,  to  destroy,  &c.   20a?  =  4a?  4-  15a? 4- 120 
Transposing  and  uniting,  a;  =120.     Ans. 

3.  A  gentleman  has  an  estate,  £  of  which  is  woodland,  |  of 
it  pasture,  and  105  acres  embrace  the  pleasure  grounds,  gar 
dens,  and  orchards.     How  many  acres  does  it  contain  ? 

ICT3"  It  is  generally  best  to  multiply  by  the  greatest  deno- 
minator first,  as  that  will  sometimes  destroy  more  than  one 
fraction.  This  is  the  case  in  this  question ;  for  6  times  ^  is 
if?  =  4a?.  Ans.  630  acres. 

4.  After  paying  away  i  and  ^  of  my  money ,  I  find  $22  yet 
in  my  purse.     How  much  had  I  at  first  ?  Ans.  $40. 

5* 


54  ALGEBRA.  [SECTION  VI. 

5.  A  man  bought  a  lot  of  ground,  for  which  he  agreed  to 
pay  as  follows :  A  of  the  mpney  on  taking  possession,  £  of  it 
in  6  months,  and  $250  at  the  end  of  the  year.  How  much 
did  he  pay  in  all  ?  .  Ans.  $600. 

6.  A  post  is  one-fourth  of  its  length  in  the  mud,  one-third 
of  it  in  the  water,  and  10  feet  above  the  water.  What  is  its 
whole  length  ?  Ans.  24  feet. 

7.  In  a  Christmas  pudding,  i  is  flour,  ^  milk,  £  eggs,  \ 
suet  and  fruit,  and  £  of  a  pound  of  spices  and  other  ingredi- 
ents.    What  is  the  weight  of  the  pudding  ? 

\CT  The  equation  will  be  #  =  £  -ff-f  £+f +}. 

y  Ans.  15  pounds^ 

Jf8.  A  lady  being  asked  what  her  age  was,  replied,  If  you 
add  -j-,  I,  and  £  of  my  age  together,  the  sum  will  be  18.  How 
old  was  she  ? 

IC7*  After  the  question  has  been  stated,  the  equation  will 
be,     £-f*-f£.  =  18.  Ans.  24  years. 

9.  What  sum  of  money  is  that,  whose  ^  part,  -£  part,  and 
i  part,  added  together,  amount  to  94  dollars  ?         Ans.  $120. 

10.  A  person  found  upon  beginning  the  study  of  his  pro- 
fession, that  he  had  passed  \  of  his  life  before  he  commenced 
his  education,  \  of  it  under  a  private  teacher,  the  same  time 
at  a  public  school,  and  four  years  at  the  university.  What 
was  his  age?  Ans.  21  years. 

11.  How  much  money  have  I  in  my  pocket,  when  the 
fourth  and  fifth  part  of  it  together,  amount  to  $2.25?  Ans.  $5. 

12.  The  3d  part  of  my  income,  said  a  person,  I  expend  ir 
board  and  lodging,  the  8th  part  of  it  in  clothes  and  washing, 
the  10th  part  of  it  in  incidental  expenses,  and  yet  I  save  $318 
a  year.     What  was  his  yearly  income  ?  Ans.  $720. 

13.  A  gentleman  bequeaths,  in  his  will,  the  half  of  his  pro- 
perty to  his  wife,  one-sixth  part  to  each  of  his  two  sons,  the 
twelfth  part  to  his  sister,  and  the  remaining  $600  to  his  servant. 
What  was  the  amount  of  his  property?  Ans.  $7200 : 


§  88.J  FRACTIONS.  55 

14.  Of  a  piece  of  metal,  -|  plus  24  ounces  is  brass,  and  |. 
minus  42  ounces  is  copper.    What  is  the  weight  of  the  piece  ? 

jCT*  The  equation,  when  formed,  will  be 

#  =  *-}- 24+ J— 42.  Ans.  216  oz. 

15.  A  farmer  mixes  a  quantity  of  grain,  so  that  20  bushels 
less  than  £  of  it  is  barley,  and  36  bushels  more  than  £  of  it  is 
oats.  How  many  bushels  are  there  in  the  whole ;  and  how 
many  of  each  sort  ? 

|C?"  In  stating  the  question;  | — 20=  barley,  and  f +  36 
r=  oats.     Ans.  96  bushels  in  all;  28  of  barley,  and  68  of  oats. 

16.  A  teacher  being  asked  how  many  scholars  he  had,  re* 
plied,  If  I  had  as  many  more,  half  as  many  more,  and  quarter 
as  many  more,  I  should  have  88.     How  many  had  he  ? 

%CT°  In  stating  the  question,  he  has  x ;  and  as  many  more 
is  another  x ;  &c.  •     Ans.  32. 

17.  In  a  mixture  of  copper,  tin,  and  lead;  16  pounds  less 
than  \  was  copper,  12  pounds  less  than  \  was  tin,  and  4  pounds 
more  than  \  was  lead.  What  was  the  weight  of  the  whole 
mixture  ;  and  also  of  each  kind  ? 

Ans.  2881b.;  and  also  1281b.,  841b.,  and  761b 

18.  What  is  that  number  whose  \  part  exceeds  its  \  part, 
by  12? 

10°  To  find  what  \  of  it  exceeds  \  of  it,  subtract  \  of  it 
from  \  of  it.     The  remainder  is  12.  Ans.  144. 

19.  What  number  is  that  whose  \  part  exceeds  its  \  part 
by  72?  Ans.  540. 

20.  A  certain  sum  of  money  is  to  be  divided  amongst  three 
persons,  A,  B,  and  C,  as  follows :  A  is  to  receive  $3000  less 
than  half  of  it,  B  $1000  less  than  the  third  of  it,  and  C  $800 
more  than  the  fourth  of  it.  What  is  the  sum  to  be  divided; 
and  what  does  each  receive  ? 

Ans.  $38400;  and  also,  $16200,  $11800,  $10400. 

21.  A  man  driving  his  geese  to  market,  was  met  by  an- 
other, who  said,  Good  morrow,  master,  with  your  hundred 


56  ALGEBRA.  [SECTION  VI. 

geese.  He  replied,  I  have  not  a  hundred ;  but  if  I  had  as 
many  more,  and  half  as  many  more,  and  two  geese  and  a  half, 
I  should  have  a  hundred.   How  many  had  he  ?    Ans.  39  geese. 

22.  A  shepherd  being  asked  how  many  sheep  he  had,  re- 
plied, If  I  had  as  many  more,  half  as  many  more,  and  7  sheep 
and  a  half,  I  should  have  just  500.    How  many  sheep  had  he? 

Ans.  197  sheep. 

23.  If  the  half,  third,  and  fourth  parts  of  my  number,  be 
added  together,  the  sum  will  be  one  more  than  my  number. 
Now,  what  is  my  number?  Ans.  12. 

24.  A  says  to  B,  Your  age  is  twice  and  §  of  my  age ;  and 
the  sum  of  our  ages  is  54  years.     What  is  the  age  of  each  ? 

Ans.  A's,  15  years;  B's,  39  years. 

25.  A  young  gentleman  having  received  a  fortune,  spent  i 
of  it  the  first  year,  i  of  it  the  second,  and  \  of  it  the  third, 
when  he  had  $2600  left.     What  was  his  whole  fortune  ? 

IC7"  In  stating  the  question,  what  was  spent,  =  the  whole 
minus  $2600.  Ans.  $12000. 

26.  A  father  leaves  four  sons,  who  share  his  property  in 
the  following  manner :  the  first  takes  half,  minus  $3000 ;  the 
second  takes  a  third,  minus  $1000  ;  the  third  takes  exactly  a 
fourth;  and  the  fourth  son  takes  a  fifth  and  $600.  What  was 
the  whole  fortune,  and  what  did  each  son  receive  ? 

Ans.  The  whole  fortune  was  $12000  ;  and  each  son 
received  $3000. 
In  generalization,  see  page  151. 


§  91.]  FRACTIONS  OF  COMPOUND  QUANTITIES.  57 

SECTION  VII. 

FRACTIONS  OF  COMPOUND  QUANTITIES 

§  89.  We  have  found,  §  72,  that  the  algebraical  method  of 
dividing,  is  to  write  the  divisor  under  the  dividend,  with  a 
straight  line  between  them.     It  is  plain  that  compound  quan- 
tities may  be  divided  in  this  way,  as  well  as  simple  quantities. 
Thus,  14 +a?  is  divided  by  3  as  follows  : 
14-fa? 
3 
§  90.  With  the  same  reason,  we  find  the  fraction  of  a  com- 
pound number,  by  multiplying  it  by  the  numerator,  and  writing 
the  denominator  under  the  product.     Thus, 

2a?— 10 


I  of  a? — 5  is 


:* 


which  is  read  2a? — 10,  both  divided  by  3. 

EXAMPLES. 

1.  What  is  |  of  8 -fa?  ?  Ans*  2±±!£. 

4 

2.  What  is  f  of  #—27  ? 

3.  What  is  f  of  3a?— 14? 

4.  What  is  4  of  9  +  5a?? 

5.  What  is  4- of  7a?— 19? 

6.  What  is  $.  of  9a?— 27  ? 

§  91.  This  may  be  changed  into  whole  numbers  by  §78. 

45.r— 135 
Thus,  q =  5a?— 15. 

7.  What  is  ^  of  86— 2a?  ? 


8.  What  is  3  of  14— 5a?? 

9.  What  is  ^T  of  2a?— 7  ? 


68  ALGEBRA.  [SECTION  VII. 


EQUATIONS.  — SECTION  7. 

PROBLEMS. 

1.  A  young  gentleman  being  asked -his  age,  said,  It  is  such 
that  if  you  add  8  years  to  it,  and  then  divide  by  3,  the  quo- 
tient will  be  9.     How  old  was  he  ? 

Stating  the  question,  x  =  his  age. 

x+8  =  with  8  added. 

x+8 

— -—  =  quotient. 

M 

Forming  the  equation,  _ J_  —  9 

o 

Multiplying  by  3,  x+8  =  27 

Transposing  and  uniting,  x=  19  the  Ans. 

2.  A  man  being  asked  what  he  gave  for  his  horse,  replied, 
that  if  he  had  given  $12  more,  a  of  the  sum  would  be  84. 
What  was  the  price  of  the  horse  ? 

Stating  tke  question,  X—  the  price. 

rr-f-12  =  when  increased. 

3*+36      3    t* 

— =  I  of  the  sum. 

4 

„       .  3x+36      m 

Forming  the  equation,        — - —  =  84 

Multiplying  by  4,  3^  +  36  =  336 

Transposing  and  uniting,  Sx  =  300 

Dividing  by  3,  x  =  100  the  Ans. 

3.  What  sum  of  money  is  that,  from  which  $5  being  sub- 
tracted, two-thirds  of  the  remainder  shall  be  $40  ?    Ans.  $65. 

4.  It  is  required  to  divide  a  line  that  is  15  inches  long  into 
two  such  parts,  that  one  of  them  may  be  |  of  the  other. 

|C7*  In  stating  the  question,  the  parts  are  a?,  and  15— a?. 

Ans.  6f ,  and  8f. 

5.  It  is  required  to  find  a  number,  such  that  if  15  be  sub- 
tracted from  it,  I  of  the  remainder  shall  be  100 ?     Ans.  140. 


§  91.]  DIVISION  OF  COMPOUND  QUANTITIES.  59 

6.  Divide  the  number  46  into  two  parts,  so  that  when  one 
is  divided  by  7,  and  the  other  by  3,  the  quotients  together 
may  amount  to  10.  Ans.  28  and  18. 

7  A  person  being  askedthe  time  of  day,  answered  that  the 
time  past  from  noon  was  equal  to  £  of  the  time  to  midnight. 
What  was  the  hour  ? 

Id*  From  noon  to  midnight  is  12  hours;  the  part  of  it 
already  past  is  x.  Ans.  20  minutes  after  5. 

8.  Two  men  talking  of  their  horses,  A  says  to  B,  My  horso 
is  worth  $25  more  than  yours ;  and  |  of  the  value  of  your 
horse  is  equal  to  f  of  the  value  of  mine.  What  is  the  value 
of  each?  lO™  The  values  will  be  x,  and  #-j-25. 

Ans.  A's,  $125;  B's,  $100. 

9.  Two  persons  have  equal  sums  of  money.  One  having 
spent  $39,  and  the  other  $93,  the  last  has  but  half  as  much  as 
the  first.     How  much  had  each?  Ans.  $147. 

10.  A  man  being  asked  the  value  of  his  horse  and  chaise, 
answered  that  the  chaise  was  worth  $50  more  than  the  horse ; 
and  that  one-half  the  value  of  the  horse  was  equal  to  one-third 
the  value  of  the  chaise.     What  was  the  value  of  each  ? 

Ans.  Horse,  $100;  chaise,  $150. 

11.  What  number  is  that,  to  which  if  I  add  13,  and  from 
fa  of  the  sum  subtract  13,  the  remainder  shall  be  13  ? 

Ans.  325. 

12.  A  man  being  asked  the  value  of  his  horse  and  saddle, 
answered  that  his  horse  was  worth  $114  more  than  his  sad- 
dle, and  that  -|  of  the  value  of  his  horse  was  7  times  the  value 
of  his  saddle.     What  was  the  value  of  each  ? 

Ans.  Saddle,  $12;  horse,  $126. 

13.  A  legacy  of  $1200  was  left  between  A  and  B,  in  such 
a  manner,  that  |  of  A's  share  was  equal  to  \  of  B's.  What 
sum  did  each  receive  ?  Ans.  A,  $640 ;  B,  $560. 


(JO  ALGEBRA.  [SECTION  VII 

14.  A  person  rented  a  house  on  a  lease  of  21  years,  and 
agreed  to  do  the  repairs  when  f  of  that  part  of  the  lease  which 
had  elapsed,  should  equal  -|-  of  the  part  to  come.  How  much 
time  will  elapse  before  he  repairs  ?  Ans.  12  years. 

15.  What  number  is  that,  to  which  if  I  add  20,  and  from 
J  of  this  sum  subtract  12,  the  remainder  shall  be  10  ? 

Ans.  13. 

16.  A  person  has  a  lease  for  99  years;  and  being  asked 
how  much  of  it  was  already  expired,  he  answered  that  §  of 
the  time  past  was  equal  to  f  of  the  time  to  come.  What  time 
had  already  past  ?  -J  Ans.  54  years. 

17.  Divide  $183  between  two  men,  so  that  ^  of  what  the 
first  receives,  shall  be  equal  to  T\  of  what  the  second  receives. 
What  will  be  the  share  of  each?  Ans.  $63,  and  $120. 

18.  Bought  sheep  for  $300,  calves  for  $100,  and  pigs  for 
$25;  and  then  laid  out  £  of  the  rest  of  my  money,  which  was 
$15,  in  getting  them  home.     How  much  had  I  at  first? 

|C7*  15  is  one  member  of  the  equation.  Ans.  $443. 

19.  A  gentleman  paid  four  laborers  $136.  To  the  first  he 
paid  three  times  as  much  as  to  the  second,  wanting  $4;  to  the 
third  one-half  as  much  as  to  the  first,  and  $6  more ;  and  tc 
the  fourth  four  times  as  much  as  to  the  third,  and  $5  more. 
How  much  did  he  pay  to  each  ? 

Ans.  To  the  first,  $26 ;  second,  $10 ;  third,  $19 ; 
fourth,  $81. 

In  generalization,  see  page  151. 


§  95.]  DIVISION    OF    FRACTIONS.  61 


SECTION    VIII. 

DIVISION  OF  FRACTIONS,  AND  FRACTIONS  OF 
FRACTIONS. 

§  92.  It  was  shown  in  §  75,  that  in  multiplying  a  fraction, 

we  multiply  the  numerator  only,  and  retain  the  denominator. 

On  the  same  principle,  a  fraction  is  divided  by  dividing  its 

numerator,  and  retaining  its  denominator. 

mL       6  2       8x     i      2*    c 

Thus,  --^3  =  -.     T^4  =  -,&c. 

§  93.  But,  supposing  we  wish  to  divide  |-  by  3.  In  this 
case,  we  cannot  divide  the  numerator  2  by  3  without  a  re- 
mainder; and  therefore  we  must  look  for  some  other  princi- 
ple to  assist  us.  We  shall  find  it  in  §  85,  where  it  was  shown 
that  a  fraction  may  be  changed  to  one  with  different  terms, 
without  altering  the  value. 

§  94.  It  is  evident,  then,  that  we  have  only  to  change  the 
fraction  which  is  to  be  divided,  to  some  equivalent  fraction, 
whose  numerator  can  be  divided  by  the  divisor  without  a  re- 
mainder. Thus,  \  can  be  changed  ^r,  if,  £J,  &c. ;  each  of 
which  can  be  divided  by  3,  giving  for  the  quotient  either  ^2T, 

§  95.  The  most  convenient  equivalent  fraction  will  be  ob- 
tained by  multiplying  both  tefms  of  the  fraction  by  the  num- 
ber which  is  to  be  the  divisor.  Because  it  is  certain  that  after 
the  numerator  has  been  multiplied  by  a  number,  the  product 
can  in  return  be  divided  by  that  number. 

Tl       4  20      _  ,20      _         4 

Thus,-^5  =  --^5;and-^5,  =  _. 

Questions.  How  do  we  multiply  a  fraction?  How  then  must  we 
divide  a  fraction?  How  can  we  do  if  the  numerator  cannot  be  di- 
vided without  a  remainder?  How  can  we  obtain  the  most  convenient 
equivalent  fraction  ? 

6 


62  ALGEBRA.  [SECTION  VIII. 

§  96.  But,  by  examining  the  example  just  given,  we  find 
the  numerator  of  the  answer  to  be  the  same  as  the  numerator 
of  the  first  fraction ;  for  the  first  numerator  has  been  multi- 
plied and  then  divided  again  by  the  same  number.  The  de» 
nominator  only  is  changed ;  and  that  has  been  done  by  mul- 
tiplying the  first  denominator  by  the  number  that  was  to  be 
the  divisor. 

§  97.  Hence  the  rule  for  dividing  a  fraction.  Divide  its 
numerator  when  it  can  be  done  without  a  remainder.  But 
if  there  would  be  a  remainder;  instead  of  dividing  the  nu- 
merator, multiply  the  denominator  by  the  divisor  for  a  new 
denominator  ;  and  leave  the  numerator  as  it  is. 


2               2 
Thus,  9-5  =  -; 

7 
11 

*•** 

EXAMPLES. 

3a: 
1.  Divide  —  by  5. 

3a: 
A,,S-  2D" 

2a: 
2.  Divide  —  by  3. 

•     • 

2a? 
Ans.  T. 

4x 
3.  Divide  y  by  2. 

2a: 

Ans.  •— - - 
7 

5x 
4.  Divide  —  by  5. 
o 

x 
Ans.  -. 

^.  .,     12a: ; 
5.  Divide  — -  by  3. 
5 

4a: 
Ans.  — -. 
5 

6.  Divide  £t?  by  2. 

Ans.  — 

7.  Divide  — — -  by  4. 
5 

x—6 
ADS-  "20-  * 

Questions.  By  this  method,  how  is  the  numerator  of  the  answer 
obtained  1  What  effect  has  this  method  upon  the  numerator]  What 
then  is  the  rule  for  dividing  a  fraction? 


Ans. 

3a?- 

-47+2/ 

64        ' 

Ans. 

7a?+4 

5a?- 

-5 

21- 

-3a? 

J  98.]  DIVISION   OF    FRACTIONS. 

8.  what  is  J  of  ^Z_J±^? 

9.  What  is  4  of  ^4? 

5  07—1 

10.  What  is  |  of—-?  .lta+63- 

fCT°  In  this  example,  although  we  can  divide  the  first  term 
in  the  numerator,  we  do  not  do  it  because  we  cannot  divide 
the  last  term. 

11.  What  is  1  of  "±3?t  Ans.^41-. 

a       Sx—2y  18a?— 1 2i/ 

27— 6y   1  27— 6y 

12.  What  is  1  of  — ^-?  Ans.  0,Q  ,  Q  *  .  . 

■       31+a?— y  279 +9a?— 9y 

«n      -    ,     <,32a?+16„  4a?  +  2 

13.  What  is  4  of  — -J,     ?  Ans.  -f--. 

7       a?— 6+y  #— 6+2/ 

§  98.  We  are  now  enabled  to  find  a  fraction  of  a  fraction 
by  the  rule  in  §  74 ;  which  is  to  multiply  by  the  numerator, 
and  divide  by  the  denominator.  To  multiply  by  the  nume- 
rator is  to  multiply  the  numerators  together.  And  to  divide 
by  the  denominator,  we  have  just  shown,  is  to  multiply  the 
denominators  together. 

EXAMPLES. 


14. 

What  is  4  of  ^? 
3        6 

Ans. 

10a?      5a? 
T8"="9# 

15. 

7a? 
What  is  |  of  t±  ? 
8 

Ans. 

21a? 
~32~* 

16. 

What  is  4  of  ^-  ? 

5          7 

Ans. 

12a? 
85* 

17. 

Kr 1 

What  is  j  of             ? 

Ans. 

10a?— 2 
21 

18. 

What  is  |  of  ^4"—? 
5            4 

Ans. 

213+6a? 
20 

Question.     How  do  we  find  a  fraction  of  a  fraction  1 


64 


ALGEBRA. 


[[SECTION  VIII. 


19.  What 

20.  What 

21.  What 

22.  What 

23.  What 

24.  What 

25.  What 

26.  What 

27.  What 

28.  What 


9y 


Ans. 


24a? 


8a? 


207  +  9*/      69  +  3y 


is  -^j-  of 


a?+61 


Ans. 


lSy 


lla?+671 

_jl    f  4a?— 20  ,  36a?— 180_  18a?~ 90 

18  iv  °   41  +  6a?'     AnS*  410+^~205  +  30aV 


2a?— lit/ 


Ans. 


12y+28a? 
14a? — 77y 


6a?+7y-10,  18a?+21y-30 

-  *  of  4+3y=§S  '        AnS'  40  +  8^=20? 


is|of£^_Z_6? 
9        6+a?-y 


Ans    *£±*£=?i 
'  54+ 9a?— 9y 


.    4       3a?+5y— 8  6a?+10v— 16 

is  f  of  -5=-*.  -  ?  Ans.     •J/gL, 


27— 4a? 
.  r3a?-18, 
"*°f2^+32? 

is  s   of  JL+to"^t 
T*       10-5i/+20a? 

is  ^  nf  16*-8  +  32y* 


135— 20a? 


EQUATIONS.  — SECTION  8. 

1.  A  farmer  wishes  to  mix  116  bushels  of  provender,  con- 
sisting of  rye,  barley,  and  oats,  so  that  it  may  contain  f  as 
much  barley  as  oats,  and  i  as  much  rye  as  barley.  How 
much  of  each  must  there  be  in  the  mixture  ? 

Stating  the  question,      a?  =  oats  ;  and  S?  =  barley. 

Then,  £  of  ^  is  ff-  =  rye. 
Forming  the  equation,  a?  +  5?+£l  =  116 

Multiplying  by  14,  14a,+  10a?+5a?==  1624 

Uniting  terms,  29a?  =1624 

Dividing  by  29,  a?  =  56     the  Ans. 


§98.]  DIVISION    OF    FRACTIONS.  65 

2.  I  paid  away  a  fourth  of  my  money,  and  then  a  fifth  of 
the  remainder,  which  was  $72.  How  much  money  had  I  at 
first? 

Stating  the  question,  x  =  what  I  have. , 

•£  =  paid  first. 
x — |  =  the  remainder. 
■j — fif  =  paid  afterwards. 
Forming  the  equation,       f — ~-u  =  72 
Multiplying  by  20,  4a?— #  =  1440 

Uniting  terms,  3a?  =  1440 

Dividing  by  3,  x  =  480     the  Ans. 

3.  After  paying  away  i  of  my  money,  and  then  \  of  the 
remainder,  I  had  $72  left.     How  much  had  I  at  first  ? 

ICT"  In  stating  the  question,  the  remainder  after  the  first 
payment  was  x — f  ;  and  }  of  that  is  | — ^. 

Then,  |-|-|— 2^4-72=  all  my  money.  Ans.  $120. 

4.  A  clerk  spends  J  of  his  salary  for  his  board,  and  §  of  the 
remainder  in  clothes,  and  yet  saves  $150  a  year.  What  is 
his  yearly  salary  ?  Ans.  $1350. 

5.  Of  a  detachment  of  soldiers,  1  are  on  actual  duty,  \  of 
them  sick,  |  of  the  remainder  absent  on  leave,  and  the  rest, 
which  is  380,  have  deserted.  What  was  the  number  of  men 
in  the  detachment?  Ans.  2280  men. 

6.  A  young  man,  who  had  just  received  a  fortune,  spent  | 
of  it  the  first  year,  and  .£  of  the  remainder  the  next  year;,  when 
he  had  $1420  left.      What  was  his  fortune  ?       Ans.  $1 1360. 

7.  If  from  i  of  my  height  in  inches,  1 2  be  subtracted,  £  of 
the  remainder  will  be  2.    What  is  my  height?    Ans    5ft.  6 in. 

8.  A  Christmas  cake  was  mixed  as  follows :  \  was  sugar, 
i  butter  and  fruit,  TJ2  eggs,  and  3  pounds  more  than  half  of  all 
these  was  flour.     How  much  did  the  cake  weigh  ? 

Ans.  12  pounds. 

9.  A,  B,  and  C,  own  together  a  field  of  36  acres.     B  has  \ 

E  8» 


66  ALGEBRA.  ^SECTION  VIII. 

more  than  A,  and  C  has  £  more  than  B.    What  is  each  man's 
share  ?         ICT*  $  more,  signifies  once  and  |  as  much. 

Ans.  A,  9  acres;  B,  12:  C,  15 

10.  A  gentleman  gave  to  three  persons  98  dollars.  The 
second  received  |.  of  the  sum  given  to  the  first;  and  the  third, 
A  of  what  the  second  had.     What  did  each  receive  ? 

Ans.  $56,  $35,  and  $7. 

11.  A  gentleman  invested  f  of  his  property  in  a  canal. 
When  he  sold  out,  he  lost  f  of  the  sum  invested,  receiving 
only  $1446.  What  was  the  value  of  his  property  when  he 
began?       JO0"  His  investment  minus  his  loss,  equals  $1446. 

Ans.  5511568. 

12.  A  boy  spent  ^  of  his  money  for  fruit ;  giving  f  of 
what  he  spent  for  oranges,  and  21  cents  for  lemons.  How 
much  money  had  he  ?  Ans.  70  cents. 

13.  A  grocer  put  490  gallons  of  beer  into  three  casks ;  of 
which  the  second  held  1£  times  as  much  as  the  first,  and  the 
third  held  J  as  much  as  both  the  others.  What  did  each 
hold?  Ans.   120,  .160,  and  210  gallons. 

14.  A  gentleman  leaves  $315  to  be  divided  among  four  ser- 
vants in  the  following  manner:  B  is  to  receive  as  much  as  A, 
and  I  as  much  more ;  C  is  to  receive  as  much  as  A  and  B, 
and  J  as  much  more ;  D  is  to  receive  as  much  as  the  other 
three,  and  \  as  much  more.     What  is  the  share  of  each  ? 

Ans.   A,  $24;  B,  $36 ;  C,  $80  ;  D,  $175 
In  generalization,  see  page  155. 


§  102.]  SUBTRACTION  OF  COMPOUND  QUANTITIES.  67 

SECTION  IX. 
SUBTRACTION  OF  COMPOUND  QUANTITIES. 

§  99.  Suppose  we  wish  to  subtract  the  expression  a?-f-6 
from  y.  It  is  evident  that  we  may  first  subtract  x ;  which 
will  give  us,  y — x.  But  we  wish  to  subtract  not  only  x,  but 
6  also.  Well,  after  we  have  subtracted  x  we  will  subtract  6 
also;  and  then  the  answer  will  be,  y— x — 6.  Therefore, 
whenever  we  wish  to  subtract  a  compound  quantity  whose 
terms  are  all  positive,  we  write  them  after  the  other  quantity 
with  -\-  changed  to  — . 

§  100.  Again,  suppose  we  wish  to  subtract  the  expression 
x — 6  from  y ;  in  which  the  number  to  be  subtracted  has  — 
instead  of  +•  As  before,  we  will  first  subtract  x,  by  which 
we  obtain  y — x.  But  the  quantity  to  be  subtracted  was  6  less 
than  x;  and  we  have  therefore  subtracted  6  too  much.  We 
will  therefore  add  6  to  our  last  answer  for  the  true  remainder, 
which  will  give  us  y — x  +  6.  Here,  we  have  changed  the 
positive  x  to  — x,  and  the  —6,  to  -f  6. 

RULE    FOR  SUBTRACTING  COMPOUND  QUANTITIES. 

§  101.  Change  all  the  signs  of  the  expression  which  is 
to  be  subtracted,  the  sign  -\-  to  — ,  and  the  sign  —  to  4-  ; 
and  then  write  the  term*  offer  the  other  quantity.  It  is  to 
be  recollected  that  in  the  quantity  from  which  we  subtract, 
the  signs  are  not  altered. 

§  102.  Although  the  subtraction  is  performed  the  moment 
the  quantities  are  written  according  to  the  above  rule,  yet, 
after  that  operation,  it  is  always  expedient  to  unite  the  term* 
if  possible. 

Questions.  How  do  we  subtract  a  compound  quantity  whose  terms 
are  all  positive?  Explain  why.  How  do  we  subtract  a  negative 
quantity'?  Explain  why.  What  is  the  Rule  for  subtracting  compound 
quantities?     What  should  be  done  after  the  quantity  is  subtracted  * 


68  ALGEBRA.  [SECTION  IX. 

EXAiMPLES. 

1.  Subtract  7x-\-G+y  from  Gy — 17. 

Ans.  Changing  the  signs  of  the  quantity  we  subtract,  we 
have,  6y— 17— 7a? — 6 — y;  which  =.  5y— 23 — 7  a?. 

2.  Subtract  4?/  + 3a?— 10  from  74— x. 

Ans.  74 — x — 4y — 3a?-|-10;  which  =84 — 4a? — 4y. 

3.  Subtract  6— a?— 3y  from  7.r-f  6i/. 

Ans.  1x-\-6y — 6-\-x-\-3y,  which  equals  8a?  +  9y — 6. 

4.  From  4a?— 3i/+27,  subtract  6t/— 12  +  x. 

Ans.  3a?— 9y  +  39. 

5.  From  6-f-a? — y,  subtract  13— 9y — x.    Ans.  2x  +  8y—7 

6.  From  8a'— 2  +  3y,  subtract  3y-\- 4— 3a?.     Ans.  11a?— 6. 

7.  From  5-f-4a?,  subtract  2 — 5a?-f-4y — z. 

Ans.  9a?— 4y+z  +  3 

8.  From  6a? — 8y,  subtract — x— y  +  oQ.  Ans.  7a? — 7y—  50. 

2a?  2a? 

9.  From  14 — ,  subtract  3a?— 12.       Ans.  26— 3a?— — • 

3  3 

Ji    _         8+7a?       .  .     3a?  8  +  7a?     .     3a? 

10.  From  — -— ,  subtract  5 -+--—.         Ans.  — 5 -• 

3  5  3  5 

,,    «        m     *!l       L  lOy— 4      A        *-     2y     lOy— 4 

11.  From  7 £,  subtract  — *- — .    Ans.  7 — J £■ 

3  o  o  b 

§103.  It  has  been  shown,  §68  and  §69,  that  any  com- 
pound quantity  may  be , considered  and  operated  upon  as  a 
simple  quantity,  by  merely  drawing  a  vinculum  above  it,  or 
enclosing  it  in  a  parenthesis.  Whenever  that  compound 
quantity  is  a  fraction,  the  line  between  the  numerator  and  de- 
nominator serves  a>  a  vinculum.  Thus,  in  the  eleventh  ex- 
ample, above,  — &■- —  is  subtracted  as  a  simple  quantity,  and 
6 

Questions.  What  effect  has  ^  vinculum  upon  a  compound  quan- 
tity 1  WThat  if  the  oomponnd  quantity  is  a  fraction?  If  a  fraction 
is  to  be  subtracted,  what  sign  is  to  be  changed  1 


§  104.]  SUBTRACTION  OF  COMPOUND  QUANTITIES.  69 

is  read,  minus  the  fraction,  <fcc.  Therefore  the  sign  of  each 
term  in  it  is  not  changed ;  but  —  is  put  before  the  whole  frac- 
tion. 

,o    t>        4x~ 6       u,        3a?+7  .         4x—6      3a?+7 

12.  From  — - — ,  subtract  — - — .         Ans. 


4     '  5  4  5 

._    „        3a?+8       .  51—  a?  3a?+8      51— a? 

13.  From — - — ,  subtract — - — .         Ans.  — — . 

2  3  2  3 

..     '         7— 2x      .  21a?  — 4 

14.  from — - — ,  subtract -— x.    i 

3  10  ,         7  —  2x      21a?— 4  , 

Ans.   — —   ~W-+X. 

15.  From  x +4— 2 a?,  subtract  3a?— 6. 

16.  From  4a? — 6y,  subtract  —  x-\-y— 7. 

17.  From  8+2a?,  subtract  — t/-+-3a?. 

,Q    n        3a?— 7       .  2— 6a? 

18.  From  — -— ,  subtract  — -— . 

2  3 

7# g 

19.  From  8a?,  subtract  3a? . 

5 

20.  From  a?,  subtract  -~^- x. 

3 

2a?— *v 

21.  From  2a?— y,  subtract  — ;r-^+y» 

3-4- a? 

22.  From  3+a?,  subtract  — !— +a?. 


UNITING  FRACTIONS  OF  DIFFERENT  DENOMINATORS. 

§  104.  By  looking  at  the  answers  to  the  last  three  exam- 
ples in  §  102,  and  also  the  next  three  in  §  103,  it  will  appear 
that  we  ought  to  have  some  rule  for  uniting  their  terms.  We 
can  easily  find  one  by  applying  the  principle  explained  in 
§  85.  For  we  have  only  to  change  each  of  the  fractions  to 
an  equivalent  fraction,  so  that  they  will  all  have  one  with  an- 
other the  same  common  denominator. 


70  ALGEBRA.  [SECTION  IX. 

Thus,  the  answer  to  the  twelfth  example  under  §  103,  is 

4x-6      3a?+7  ,      „   ■ 

— — .     JNow  each  ot  these  two  fractions  maybe 

changed  to  20ths,  by  multiplying  the  first  by  5,  and  the  last 

fpu          ...   ,       '.             20a;—* 30      12ar+28     ,  .  . 
by  4.     They  will  then  become ■■ — I which  = 

(20a:— 30)  —  (12a:  -J-  28)  _  20ar— 30  —  12a;  —  28  __  8x  —  58 
90  20  20~~ 

4x  —  29 

7     io-"* 

§  105.  Thus  we  have  the  rule  for  uniting  fractions  of  dif- 
ferent denominators.  Multiply  all  the  denominators  toge- 
ther for  a  new  denominator;  and  each  numerator  by  all  the 
denominators  except  its  own,  for  new  numerators :  remem- 
bering that  if  a  compound  numerator  follows  minus  — ,  all 
the  signs  in  it  must  be  changed  the  m  ment  one  denomina- 
tor is  used  for  the  whole  quantity  ;  that  is,  when  the  short 
vinculums  are  destroyed  for  the  purpose  of  making  the  longer 
one. 

examples. 

1.  Unite  the  terms  in  the  answer  to  the  13th  sum  in  §  103. 
Operation. 
/3a- -f  8      51—  a^  _    9.r+24_  102— 2x\  _ 
V2  3~~  '™V      6  (5       / 

9a--f24  —  102-f2a?_lla;— 78 
~~ 6  6       * 

Note. — In  this  operation  the  first  minus  has  reference  to 

the  whole  quantity  /  — —  j ;  the  second  minus  to  the  whole 

.  \  no 2a?\ 

quantity  ( ).     In  this  last  quantity,  102  has  no  sign 

before  itself  and  is  therefore  positive.     Now,  when  the  line 

Questions.  How  can  fractional  terms  be  united  1  Give  the  rule. 
What  if  integers  are  multiplied  with  the  fractions  1 


§106.]  DIFFERENT    DENOMINATORS.  71 

between  the  numerator  and  denominator  is  carried  through 
the  ic'ole  quantity,  the  vinculum  of  102—2.7;  is  destroyed; 
and  then  is  the  time  for  changing  the  signs  for  subtracting. 

§  106.   Whenever  there  are  integers  to  be  united  with  frac- 
tions, they  may  be  changed  to  fractions,  by  putting  the  number 

6  x 

1  under  them  for  the  denominator.     Thus,  6=—;    #  =  -. 

2.  Unite  the  terms  in  the  quantity  7 -  — - . 

5  6 

Operation. — The  quantity  is —  — ~ — . 

15  6 

7  210 

Both  terms  of  -  multiplied  by  5  and  by  6,  equals  — — . 

Both  terms  of f-  multiplied  by  6,  equals ^. 

Both  terms  of ^ —  multiplied  by  5,  equals =^~ — . 


Therefore, 


/7      2y      lOy—  4\  _  ,2\0 _12y ^50y  —  20\ 
\l~~5~       6       /       \~30T      "30  30   ~) 

210— 12y— 50yf20     _  230-r62y         115— 31y 
30  30  15 

«    it  •       u  -i  8-J-7.T      "      Sx 

3.  Unite  the  terms  in  the  quantity, 5-- . 

3  5 

40  +  35^— 75— 9x      26.r— 35 
AnS* IS" 15— 

i  2iP 

4.  Unite  the  terms  in  the  quantity,  26  — 3#-f-— -. 

o 

^m     §x-\-2x      n„     7x 

Ans.  26 ~ —  =  26 -. 

o  o 

2x     21  x 4 

5.  Unite  the  terms  in  the  quantity,  7 — —      ln     -fa:. 

Ans. 


72  ALGEBRA.  [SECTION  IX 

6  +  2x     4x  —  3 

6.  Unite  the  terms  in  the  quantity,   12 —    —        — 

J&  Ans.  *"=£?. 

8  l0 

x     x     x 

7.  Unite  the  terms  in  the  quantity,  -+- +-- 

13#  .    # 

Ans.  -g— •+ff 

8.  Unite  the  terms  in  the  quantity,  4+  3  —  *  — ~g* 
^J*  The  fractions  may  be  united  }  Ang    j^* 

by  themselves.  S 

3*  4z 

9.  Unite  the  terms  in  the  quantity,  ±z  ——+z — jr. 

43z 

Ans.    5Z rr-« 

35 

Sy— 4 

10.  Unite  the  terms   in   the   quantity,   y - h2y— 

3  15  15 

6x+7  r         3*— 4 

11.  Subtract  — - —  from  — — . 

4  / 

5x— 6  r        S—x 

12.  Subtract  — — -   from  — - — 


x  x 

13.  Subtract  3—-  from  7-. 

2     x 

14.  Subtract  Sx -—  from  5a?. 

4 

15.  Subtract  -? — - from  - -J-. 

16.  Subtract  £^+VS£!  from  4*-f. 


§106.]  SUBTRACTION  OF  COMPOUND  QUANTITIES.  73 

EQUATIONS.  — SECTION  9. 

1.  There  are  two  numbers,  whose  sum  is  140;  and  if  4 
times  the  less  be  subtracted  from  3  times  the  greater,  the  re- 
mainder will  be  70.     What  are  the  numbers  ? 

Stating  the  question,  x  —  the  greater. 

140 — x==  the  less. 

Sx  =  3  times  the  greater. 
560— 4a:  =  4  times  the  less. 
Forming  the  equation,  3a?— 560  + 4a?  =  70 
Transposing,  uniting,  and  dividing,  x  mm  90 

Ans.  Greater  number,  90 ;  less,  50. 

2.  A  person,  after  spending  $100  more  than  a  third  of  his 
yearly  income,  found  that  the  remainder  was  $150  more  thai 
half  of  it.     What  was  his  income?  Ans.  $1500 

3.  Two  men,  A  and  B,  commenced  trade.  A  had  twice  a* 
much  money  as  B  ;  he  has  since  gained  $50,  and  B  has  losf 
$90 ;  and  now  the  difference  between  A's  and  B's  money,  is 
equal  to  three  times  what  B  has.  How  much  had  each  when 
they  commenced  trade?  Ans.  A,  $410;  B,  $205. 

4.  A  man  bought  a  horse  and  chaise  for  $341.  If  $  of  the 
price  of  the  horse  be  subtracted  from  twice  the  price  of  the 
chaise,  the  remainder  will  be  the  same  as  if  f  of  the  price  of 
the  chaise  be  subtracted  from  three  times  the  price  of  the 
horse.     What  was  the  price  of  each  ? 

IC  If  the  price  of  the  chaise  be  a?,  and  the  price  of  the  horse 

be  341— a:;  then  the  first  remainder  will  be  2a? = . 

But  when  the  fraction  is  destroyed,  the  vinculum  is  taken 
away,  and  therefore  the  last  sign  must  be  changed  from  — 
to  -j-.  Ans.  Chaise,  $189;  horse,  $152. 

5.  A  gentleman  bought  a  watch  and  chain  for  $160.  If  | 
of  the  price  of  the  watch  be  subtracted  from  six  times  the  price 
of  the  chain,  the  remainder  will  be  the  same  as  if  T5f  of  the 

7 


74  ALGEBRA.  ^SECTION  IX. 

price  df  the  chain  were  subtracted  from  twice  the  price  of  the 
watch.     What  was  the  price  of  each  ? 

Ans.  Watch,  $112;  chain,  $48. 

6.  Divide  the  number  204  into  two  such  parts,  that  if  T  of 
the  less  were  subtracted  from  the  greater,  the  remainder  will 
be  equal  to  ^  of  the  greater  subtracted  from  four  times  the 
less.  Ans.  Greater,  154;  less,  50. 

7.  Two  travellers,  A  and  B,  found  a  purse  of  money.  A 
first  takes  out  $2  and  *  of  what  remains ;  and  then  B  takes 
$3  and  j-  of  what  remains  ;  and  it  is  found  that  each  has  the 
same  sum.    How  much  money  was  in  the  purse  ?     Ans.  $20. 

8.  A  shepherd  was  met  by  a  band  of  robbers,  who  plun- 
dered him  of  half  of  his  flock  and  half  a  sheep  over.  After- 
wards a  second  party  met  him,  and  took  half  of  what  he  had 
left,  and  half  a  sheep  over;  and  soon  after  this,  a  third  party 
met  him,  and  treated  him  in  the  like  manner;  and  then  he  had 
5  sheep  left.     How  many  had  he  at  first?        Ans.  47  sheep. 

9.  A  gentleman  hired  a  laborer  for  20  days,  on  condition 
that  for  every  day  he  worked  he  should  receive  14  shillings; 
but  for  every  day  he  was  idle  he  should  forfeit  6  shillings. 
At  the  end  of  the  20  days  he  received  160  shillings.  How 
many  days  did  he  work,  and  how  many  days  was  he  idle  ? 

Ans.  He  worked  14  days,  and  was  idle  6. 

10.  Divide  the  number  48  into  two  such  parts,  that  the  ex- 
cess of  one  of  them  above  20,  shall  be  three  times  as  much  as 
the  other  wants  of  20. 

f£J*  The  excess  of  a  number  above  20  is  obtained  by  sub- 
tracting 20  from  it.  Ans.  32  and  16. 

11.  A  person  in  play  lost  a  fourth  of  his  money,  and  then 
won  back  3*. ;  after  which  he  lost  a  third  of  what  he  now 
had,  and  then  won  back  2s. ;  lastly,  he  lost  a  seventh  of 
what  he  then  had,  and  then  found  he  had  but  12s.  remaining. 
What  had  he  at  first?  Ans.  20s. 

In  generalization,  see  page  158. 


§   109.]  RATIO    AND    PROPORTION.  75 

SECTION    X. 
RATIO  AND  PROPORTION. 

§  107.  When  an  unknown  quantity  is  not,  either  by  itself, 
or  in  some  connexion  with  others,  known  to  be  equal  to  some 
known  quantity  or  set  of  quantities  ;  we  may  sometimes  find 
that  there  is  a  comparison  between  it  and  some  known  quan- 
tity, which  is  the  same  as  the  comparison  between  two  known 
quantities. 

Thus,  suppose  I  buy  27  yards  of  cloth  for  $72,  and  wish 
to  sell  for  $16  so  much  of  it  as  cost  me  $16.  In  this  case 
the  number  of  yards  to  be  sold  is  not  equal  to  any  other 
quantity  that  is  mentioned.  But  we  suppose  that  it  must 
compare  with  the  number  of  yards  bought,  in  the  same  man- 
ner that  $16  compares  with  $72.  By  knowing  this  compari- 
son, we  can  find  the  number  of  yards ;  because,  as  $16  is  | 
of  $72,  so  the  number  of  yards  to  be  sold  must  be  |  of  the 
number  of  yards  bought.     It  is  6  yards. 

§  108.  It  will  be  seen  that  the  comparison  in  this  example 
consists  in  observing  how  many  times  one  of  the  numbers  is 
contained  in  the  other.  72  is  contained  in  16,  two-ninths  of 
a  time.  When  a  comparison  of  this  kind  is  made,  the  result 
that  is  obtained  is  called  their  ratio.  Thus,  in  comparing 
the  numbers  3  and  4,  we  find  that  4  is  contained  in  3,  three- 
fourths  of  a  time ;  and  therefore  we  say  the  ratio  of  3  to  4  is  -J. 

§  109.  The  pupil  must  remember  that  the  ratio  of  one 
number  to  another,  always  signifies  how  the  first  number 
compares  with  the  last.  Thus,  the  ratio  of  8  to  5,  is  } ;  that  is, 
8  is  |  of  5.  Hence  the  ratio  is  expressed  by  making  the  first 
term  to  be  a  numerator,  and  the  last  to  be  the  denominator. 

Questions.  Do  we  ever  make  use  of  comparison  in  algebra?  Ex- 
plain how.  In  what  does  the  comparison  consist]  What  is  that 
kind  of  comparison  called  in  mathematical  language  ?  How  can  a 
ratio  be  expressed  ?     Why  1 


76  ALGEBRA.  [SECTION  X. 

§  1 10.  In  the  example  just  furnished  relative  to  the  cloth; 
the  ratio  of  the  money  paid,  to  the  money  obtained  for  a  part 
of  the  cloth;  (that  is,  the  ratio  of  $72  to  $i6,)  is  T*«*§.  And 
so  also  the  ratio  of  the  cloth  bought,  to  the  cloth  sold;  (that  is, 
the  ratio  of  27  yards  to  6  yards,)  is  2?  which  equals  2.  Here 
we  see,  that  although  the  ratios  are  differently  expressed,  they 
are,  notwithstanding,  equal  to  one  another. 

§111.  When  the  ratio  of  two  quantities  is  equal  to  the 
ratio  of  other  two  quantities,  there  is  said  to  be  a  proportion 
between  them;  that  is,  an  equality  f  ratios  is  called  a  pro- 
portion. 

§  1 12.  Our  chief  business  with  ratios  at  present,  is  to  learn 
when  they  form  a  proportion ;  that  is,  when  they  are  equal  to 
one  another.  Now,  as  they  may  be  expressed  in  the  form  of 
a  fraction;  it  is  evident,  that  when  they  are  brought  to  a  com- 
mon denominator,  if  the  fractions  are  equal  their  numerators 
will  be  the  same,  and  if  they  are  not  equal  their  numerators 
will  not  be  the  same. 

For  example,  is  1 1  to  21  =  33  to  63  ?  We  pursue  our  in- 
quiry as  follows :  11  to  21  is  the  same  as  ^j,  and  33  to  63  is 
the  same  as  j4«  We  bring  the  fractions  to  a  common  deno- 
minator by  §  105. 

11       ,33       693       „    693 

—  and  —  =    and . 

21  63      1323  1323 

We  find  they  are  equal,  and  the  four  terms  11  to  21  =33  to 
63  are  proportional. 

§  113.  Although  ratios  are  sometimes  expressed  fraction- 
ally, they  are  generally  expressed  as  follows:  11  :  21  and 
33  :  63 ;  that  is,  11  divided  by  21,  33  divided  by  63.  The 
pupil  will  see  that  we  employ  the  same  sign  that  expresses 
division,  with  the  exception  of  the  —  between  the  two  dots 


Question.  ■  Can  a  ratio  be  equal  to  another,  and  yet  be  differently 
expressed  ?  Give  an  example.  What  do  we  call  an  equality  of 
ratios]  How  may  we  determine  whether  ratios  are  equal?  Give 
an  example.   Are  ratios  always  expressed  by  fractions  ?   How  else  1 


§117.J  RATIO    AND    PROPORTION.  77 

The  sign    :    is  read  is    to,  and   the  foregoing  examples  are 
read  11  is  to  21  and  33  is  to  63. 

§114.  When  four  quantities  are  proportional,  they  are 
written  thus,  11  :  21  : :  33  :  63.  The  sign  ::  is  read  as; 
and  the  whole  expression  is  read,  11  is  to  21  as  33  is  to  63. 

§115.  In  a  proportion,  the  first  and  the  last  terms  are 
called  extremes,  and  the  two  middle  terms  are  called  means. 
In  the  above  proportion,  11  and  63  are  the  extremes,  and  21 
and  33  are  the  means. 

§  116.  In  order  to  derive  any  important  use  from  a  pro- 
portion, we  wish  the  pupil  to  recollect  the  method  employed 
to  find  whether  four  quantities  are  proportional.  We  multi- 
plied (see  §112;  ihe  first  numerator  by  the  last  denominator, 
to  find  one  new  numerator.  These  were  the  two  extremes. 
We  also  multiplied  the  last  numerator  by  the  first  denomina- 
tor, to  find  the  ther  new  numerator.  These  were  the  two 
means.  And  hence  we  learn,  that  if  four  quantities  are  pro- 
portional, the  product  of  the  two  extremes  is  equal  to  the 
product  of  the  means. 

RULE. 

§  117.  «/?  proportion  may  be  reduced  to  an  equation  by 
multiplying  the  extremes  together  for  one  member ;  and 
multiplying  the  means  t  gether  for  the  other  member. 
Thus,  2:7::  8  :  x,  becomes  in  an  equation  2x  =  56  ; 
whence    x  =  28. 

Or,  the  fourth  term  may  befund  by  multiplying  the  two 
means  t>  gether,  and  dividing  their  product  by  the  first 
extreme. 


Questions.  How  is  a  proportion  written  ?  Which  terms  are  the 
extremes  ?  Which  terms  are  the  means  ?  In  hrinjjintr  the  n.ti  s  to 
a  common  denominator,  what  did  we  do  with  the  extremes]  Y\  hat 
did  we  do  with  the  two  means?  What  principle  does  this  show  1 
What  then  can  we  do  with  a  proportion  ? 


7* 


78  ALGEBRA.  [SECTION  X. 


EQUATIONS.— SECTION  10. 

1.  If  you  divide  $75  between  two  men  in  the  proportion  of 
3  to  2,  what  will  each  man  receive  ? 

Stating  the  question,  x  =  the  share  of  one. 

75  —  x  =  the  share  of  the  other. 
Making  the  proportion.  x  :  75 — x  : :  3  : 2 


2x  =  225— 3a? 


Multiplying  ext.  and  means 
to  reduce  to  an  equation, 
Transposing  and  uniting,  5x  =  225 

Dividing,  x  =  45 

Ans.  $45;  and  $30. 

2.  Divide  $150  into  two  parts,  so  that  the  smaller  may 
be  to  the  greater  as  7  to  8.  Ans.  70 ;  and  80. 

3.  Divide  $1235  between  A  and  B,  so  that  A's  share  may 
be  to  B's  as  3  to  2.  Ans.  A's  share  $741  ;  B's  $494. 

4.  Two  persons  buy  a  ship  for  $8640.  Now,  the  sum 
paid  by  A  is  to  that  paid  by  B,  as  9  to  7.  What  sum  did 
each  contribute  ?  Ans.  A  paid  $4860  ;  B  $3780. 

5.  A  prize  of  $2000  was  divided  between  two  persons, 
whose  shares  were  in  proportion  as  7  to  9.  What  was  the 
share  of  each?  Ans.  $875;  and  $1125. 

6.  A  gentleman  is  now  30  years  old,  and  his  youngest 
brother  20.     In  how  many  years  will  their  ages  be  as  5  to  4? 

iO0"  After  stating  the  question,  the  proportion  will  be 
30+a?:20+#::  5:4.  Ans.  20  years. 

7.  What  numbei  is  that,  which,  when  added  to  24,  and 
also  to  36,  will  produce  sums  that  will  be  to  each  other  as  7 
to  9?  Ans.   18. 

8.  Two  men  commenced  trade  together.     The  first  put  in 


§  118.]  RATIO    AND    PROPORTION.  79 

$40  more  than  the  second  ;  and  the  stock  of  tne  first  was  to 
that  of  the  second  as  5  to  4.     What  was  the  stock  of  each  ? 

Ans.  $200;  and' $160. 

9.  A  gentleman  hired  a  servant  for  $100  a  year,  together 
with  a  suit  of  clothes  which  he  was  to  have  immediately.  At 
the  end  of  8  months,  the  servant  went  away,  and  received 
$60  and  kept  the  suit  of  clothes.  What  was  the  value  of  the 
suit  of  clothes  ?  Ans.  $20. 

10.  A  ship  and  a  boat  are  descending  a  river  at  the  same 
time ;  and  when  the  ship  is  opposite  a  certain  fort,  the  boat 
is  13  miles  ahead.  The  ship  is  sailing  at  the  rate  of  5  miles, 
while  the  boat  is  going  3.  At  what  distance  below  the  fort 
will  they  be  together? 

The  ship  sails  x  miles  from  the  fort ;  the  boat  will 
sail  13  miles  less.  Ans.  32£  miles. 

§118.  It  is  very  often  the  case  that  a  problem  is  easily 
solved  by  using  simply  the  ratio,  instead  of  a  proportion. 

Operation  by  Ratio. 

%C3*  In  these  questions  the  pupil  must  not  use  any  pro- 
portions. 

11.  Divide  40  apples  between  two  boys  in  the  proportion 
of  3  to  2. 

Stating  the  question,  x  =  the  share  of  one. 

Now,  as  the  ratio  of  the  first  to  the  second  is  | ;  then  the 
ratio  of  the  second  to  the  first  is  J.     Therefore, 

Org  « 

—  =  the  share  of  the  second. 
o 

2<£ 

Forming  the  equation,  a?-f-— =  40 

o 

Multiplying  by  3,  3#-f  2x  =  120 

Uniting  terms,  x  =  24 

Ans.  24,  and  16. 


80  ALGEBRA.  [SECTION  X. 

12.  Three  men  trading  in  company,  gain  $780.  As  often 
as  A  put  in  $2,  B  put  in  $3,  and  C  put  in  $5.  What  part  of 
the  gain  must  each  of  them  receive  ? 

Stating  the  question,  x  =  A's  share. 

—  =  B's  share. 
2 

—  =  C's  share. 
z 

Sx      5x 

Forming  the  equation,  x-\-  —  -+-  -7-  =  780. 

Ans.  A,  $156;  B,  $234 ;  C,  $390. 

13.  Two  butchers  bought  a  calf  for  40  shillings,  of  which 
the  part  paid  by  A,  was  to  the  part  paid  by  B,  as  3  to  5. 
What  sum  did  each  pay  ?  Ans.  A  paid  15*. ;  B,  25s. 

14.  Divide  560  into  two  such  parts,  that  one  part  may  be 
to  the  other  as  5  to  2.  Ans.  400,  and  160. 

15.  A  field  of  864  acres  is  to  be  divided  among  three  farmers, 
A,  B,  and  C ;  so  that  A's  part  shall  be  to  B's  as  5  to  1 1,  and 
C  may  receive  as  much  as  A  and  B  together.  How  much 
must  each  receive?         Ans.  A,  135;  B,  297;  C,  432  acres. 

16.  Three  men  trading  in  company,  put  in  money  in  the 
following  proportion ;  the  first  3  dollars  as  often  as  the  second 
7,  and  the  third  5.  They  gain  $960.  What  is  each  man's 
share  of  the  gain  ?  Ans.  $192;  $448;  $320. 

17.  Find  two  numbers  in  the  proportion  of  2  to  1,  so  that 
if  4  be  added  to  each,  the  two  sums  will  be  in  proportion  of 
3  to  2. 

IC7*  The  last  expression  means  that  the  greatest  is  |  of  the 
smallest ;  or  the  smallest  is  -|  of  the  greatest.      Ans.  8  and  4. 

18.  Two  numbers  are  to  each  other  as  2  to  3  ;  but  if  50 
be  subtracted  from  each,  one  will  be  one-half  of  the  other. 
What  are  the  numbers?  Ans.    100  and  150. 

19.  A  sum  of  money  is  to  be  divided  between  two  persons, 
A  and  B  ;  so  that  as  often  as  A  takes  $9,  B  takes  $4.     Now 


§  118.]  RATIO    AND    PROPORTION.  81 

it  happens  that  A  receives  $15  more  than  B.     What  is  the 
share  of  each  ?  Ans.  A,  $27  ;  B,  $12. 

20.  There  are  two  numbers  in  proportion  of  3  to  4 ;  but  if 
24  be  added  to  each  of  them,  the  two  sums  will  be  in  the  pro- 
portion of  4  to  5.    What  are  the  numbers  ?     Ans.  72  and  96. 

21.  A  man's  age  when  he  was  married  was  to  that  of  his 
wife  as  3  to  2 ;  and  when  they  had  lived  together  4  years,  his 
age  was  to  hers  as  7  to  5.  What  were  their  ages  when  they 
were  married?  Ans.  His  age,  24;  hers,  16  years. 

22.  A  certain  man  found  when  he  married,  that  his  age  was 
to  that  of  his  wife  as  7  to  5.  If  they  had  been  married  8  years, 
sooner,  his  age  would  have  been  to  hers  as  3  to  2.  What 
were  their  ages  at  the  time  of  their  marriage  ? 

Ans.  His  age,  56  years ;  hers,  40. 

23.  A  man's  age,  when  he  was  married,  was  to  that  of  his 
wife  as  6  to  5 ;  and  after  they  had  been  married  8  years,  her 
age  was  to  his  as  7  to  8.  What  were  their  ages  when  they 
were  married  ?  Ans.  Man,  24  ;  wife,  20  years. 

24.  A  bankrupt  leaves  $8400  to  be  divided  among  four 
creditors,  A,  B,  C,  and  D,  in  proportion  to  their  claims. 
Now,  A's  claim  is  to  B's  as  2  to  3 ;  B's  claim  to  C's  as  4  to 
5 ;  and  C's  claim  to  D's  as  6  to  7.  How  much  must  each 
creditor  receive  ? 

Ans.  A,  $1280;  B,  $1920 ;  C,  $2400;  D,  $2800. 

25.  A  sum  of  money  was  divided  between  two  persons,  A 
and  B,  so  that  the  share  of  A  was  to  that  of  B  as  5  to  3.  Now, 
A's  share  exceeded  f  of  the  whole  sum  by  $50.  What  was 
the  share  of  each  ?  Ans.  $450,  and  $270. 

In  generalization,  see  page  154 


82  ALGEBRA.  [SECTION  XI 


SECTION  XI. 

EQUATIONS    WITH    TWO    UNKNOWN 
QUANTITIES. 

§  119.  It  frequently  happens,  that  several  unknown  quan- 
tities are  introduced  into  a  problem.  But  when  this  is  the 
case,  if  the  conditions  will  give  rise  to  as  many  equations, 
independent  of  each  other,  as  there  are  unknown  quantities, 
there  is  no  difficulty  in  finding  the  value  of  each  quantity. 

§  120,  An  equation  is  said  to  be  independent  of  another 
when  it  cannot,  either  by  multiplication  or  by  division,  be 
changed  into  that  other,  Thus,  Ix — i/=47,  is  independent  of 
the  equation  10?/-f  4x  =  50 ;  because  one  of  them  cannot  be  so 
altered  as  to  make  the  other.  But,  Ix — 1/=47,  is  not  inde- 
pendent of  the  equation  2\x — 3i/=141  ;  because  the  last  is 
made  by  multiplying  the  first  by  3. 

§  121.  At  present  we  will  attend  to  those  equations  only 
that  include  two  unknown  quantities,  each  represented  by  a 
different  letter  from  the  other. 

§  122.  In  equations  that  contain  two  unknown  quantities, 
our  first  object  must  be  to  find  the  value  of  one  of  them ;  and 
in  order  to  do  this,  the  preliminary  step  is  to  derive  from  the 
equations  that  are  given,  another  equation  which  shall  have 
but  one  unknown  quantity.  This  operation  is  called  elimi- 
nating, or  exterminating  the  other  unknown  quantities. 

§  123.  There  are  three  different  methods  of  forming  one 
equation  with  one  unknown  quantity  from  two  equations 
containing  two  unknown  quantities.  With  each  of  these,  the 
learner  should  become  familiar;  as  it  is  sometimes  convenient 
to  use  one  of  them,  and  sometimes  another. 

Questions.  When  is  one  equation  said  to  be  independent  of  an- 
other1? Explain.  When  there  are  several  unknown  quantities,  how 
many  independent  equations  are  necessary  to  solve  the  question? 
What  is  it  to  eliminate  or  exterminate  an  unknown  quantity] 


$   125. J      EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  83 

FIRST  METHOD  OF  EXTERMINATION. 

§  124.  It  is  necessary  here  to  recollect  what"  was  stated  in 
$  50  and  §  53,  that  when  equals  are  added  to  equals,  their 
sums  will  be  equal ;  and  also,  when  equals  are  st'btracted 
from  equals,  the  remainders  are  equal. 

Thus,  suppose  we  have  the  equation    a?-j-14  =  36, 
and  suppose  also  that  we  know  that  y  =  8  ; 

then  if  we  will  add  the  two  first  members  together,  and  also 
the  two  last,  the  members  will  still  be  equal  to  one  another,  as 
follows:     a?+ 14+2/=  36 -f8. 

And  also  if  we  subtract  y  from  the  first  member,  and  8  from 
the  second,  the  members  will  still  be  equal  to  one  another ; 
thus,  #-f  14— ?/  =  36— 8. 

§  125.  This  principle  can  be  easily  applied  for  the  exter- 
mination of  unknown  quantities.  For,  if  in  both  of  two  equa- 
tions, one  of  the  unknown  quantities  has  the  same  co-efficient, 
but  after  different  signs;  it  is  evident  that  if  we  add  both  equa- 
tions together,  viz.  the  first  member  to  the  first  member,  and 
the  last  member  to  the  last  member;  and  then  unite  temia, 
we  shall  cancel  the  two  quantities  that  are  alike  with  different 
signs;  a  new  equation  will  be  formed,  in  which  that  unknot* 
quantity  will  disappear. 

EXAMPLES. 

1.  Given  the  two  C  3#-f-2i/=26  >  to  find  the  values  of 
equations,  £  5x — 2y  =38  }    x  and  y. 

Adding  together  the  two  right  hand  members,  and  also  th# 
two  left ;  we  have  the  equation 

3.r  +  2y  -f-  5x— 2y  =  26  -f  38 
Uniting  terms  and  canceling  2y,     Sx  =  64 
Dividing,  x  =  S 


Questions.  What  is  the  effect  of  adding  two  equations  together, 
(the  first  member  to  the  first  member,  and  the  last  member  to  the 
last  member)  1  In  what  case  can  an  unknown  quantity  be  extermi- 
nated by  adding  two  equations  together? 


84  ALGEBRA.  [SECTION  XI 

Now,  if  x  =  8 ;  then  in  the  first  equation,  how  much  will 
3x  equal  ? 

The  first  equation  will  then  become     24-J-2y  =  26, 
from  which  we  may  find  the  value  y  =  l. 

_.         .  .         C7a?-r-4u  =  58  >  to  find  the* values 

2.  Given  the  equations,  1 9x_4*  =  38  $  of  x  and  y. 

Ans.  a?  =  6. 
Question.   Why  do  you  add  the  equations  ? 
If  a?  =  6,  what  does  7x  in  the  first  equation  equal  ?     Then 
what  does  y  equal  ?  Ans.  y  =  4. 

_     ~.         ,  C  5a?-l-6v  =  58  ?  to  find  the  values 

3.  Given  the  equations,  J  ^  6*  =  34  $  of  *  and  y, 

fCT"  In  this  example,  it  is  plain  that  we  cannot  destroy  the 
y's  by  adding  them  together.  But  we  have  before  seen,  §  62, 
that  if  all  the  signs  are  changed,  the  equation  will  not  be  af- 
fected. Let  us  then  change  the  signs  of  the  second  equation, 
so  that  the  y's  may  have  different  signs.     The  two  equations 

will  then  become    j  2a?— 61/ ==—34  \    wn*cn' 

when  added  together,  become  5x+6y— 2a?—  6y  =  58— 34 
Uniting  terms,  -    3a?  =  24 

Therefore,  x  =  8. 

Question.   What  operation  is  performed  by  changing  signs  ? 

§  126.  In  the  last  example,  if  we  take  the  equations  before 

the  alteration  of  the  second,  thus,    2„   Ta   —q^c    an(^  sud" 

tract  the  second  from  the  first,  the -result  will  be  the  same  as 
it  was  by  changing  the  signs  and  adding.     As  follows : 
5x+6y—2x—6y  =  58—34. 

Whence  we  learn  that,  if  in  both  equations  one  of  the  un- 
known quantities  has  the  same  co-efficient  and  also  the  same 
sign  ;  and  we  subtract  one  equation  from  the  other,  (viz.  the 
first  member  from  the  first  member,  and  the  second  member 

Questions.  In  what  case  can  an  unknown  quantity  be  extermi- 
nated by  subtracting  :>ne  equation  from  another?   Explain  the  reason. 


5a?+6t/=64 

2x +6?/==  58 
__  _ 


§  127.]       EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  85 

from  the  second  member,)  and  unite  the  terms;  we  shall  form 
a  new  equation  in  which  that  unknown  quantity  will  disappear. 

§  1 2  7.  It  is  evident  that  we  may  suppose  the  signs  changed ; 
and  so  unite  the  terms  immediately,  without  actually  writing 
the  whole  work.     This  must  be  done  hereafter. 

.     ~.        A,  C6#  +  7w  =  79?  to  find  the  values 

4.  Given  the  equations,  £  te^,B1^  of  x  and  y. 

Subtracting  second  from     )  .    2ft 

first,  and  unite.  3 

Ans.  a?  =  5;  y  =  7. 

5.  Given    J  2x+§l58  J    t0  find  *  and  tf 
ICT3"  In  this  example,  the  y's  are  alike  in 

both  equations ;  and  are  therefore  called  iden- 
tical terms.  As  they  have  the  same  sign,  in 
order  to  cancel  them  we  subtract  the  second 
from  the  first,  and  unite.  Ans.  x  =  2 ;  y  =  9. 

6-Given    lltZZlll]    ">  find  ^  and  y. 
Which  are  the  identical  terms  in  this  example  ? 
ICT*  To  cancel  the  #'s,  subtract  the  upper  from  the  lower. 

Ans.  x  =  40;  i/==ll. 

_,    ~;  C  12#-f  8y=92  } 

7.  Given    |  V2xZ2ly  =  63  $    t0  find  *  and  ^ 

Ans.  x  =  7;  y  =  1. 

8.  Given    J  3^g=^  J    to  find  X  and  y. 

ICT3*  As  the  signs  of  the  identical  terms  are  2a?-{-2i/=18 
unlike,  to  cancel  them  we  add  the  second  to  33? — 2y=  7 
the  first,  and  unite.  5a?         =25 

Ans.  x  =  5;  y  =  4. 

9.  Given   J  _^+:g=  -12  \    t0  find  *  and  * 

Ans.  a?  =  4;  i/==  2. 

Questions.  Is  it  necessary  to  write  out  the  whole  work  ?  What 
are  identical  terms  ? 

8 


86  ALGEBRA.  [SECTION  XI 

10.  Given    5       To?"  %a  £    to  find  #  and  ty. 

I    x-\-'Zy  =  14  ^  J 

|C7^  In  this  example,  neither  of  the  unknown  quantities 
has  the  same  co-efficient  in  both  equations.  But  both  mem- 
bers of  the  last  equation  can  be  multiplied  by  3,  without  de- 
stroying the  equality,  §  82 ;  and  then  the  co-efficients  of  the 

a?'s  will  be  alike  in  both  equations.     Thus,   J  _    Ifi^— 4-2  C 

Ans.  x  =  10;  y  =  2. 

11.  Given    ^  3^^^l8g  \    to  find  x  and  y. 

|C7"  Multiply  the  second  by  2.  Ans.  x  =  8  ;  y  —  16. 

12.  Given    j  3™  _  ^—  2  C    to  **nd  x  and  &' 

IC7*  Make  t/'s  identical.  Ans.  x  =  3  ;  y  =  7. 

13.  Given    ^  4^  +  3y  =  22  5    t0  find  x  and  ^ 

|C7*  Multiply  the  first  by  2  and  the  ;r's  will  be  identical. 

Ans.  x  =  4  ;  1/  =b  2 

14.  Given    ^+5zls2  £    t0  find  *  and  *< 

Ans.  rr  =  7;  2-  =  8 

15.  Given    \fx^yZ^\    to  find  z  and  y. 

Ans.  a?  =  4;  y  =  5. 

16.  Given    S^lg^i}?    to  find  x  and  y. 

Ans.  a:  =  8  ;  y  =  5. 

17.  Given    J  4£  +  *  "  ^  £    to  6nd  y  and  * 

Ans.  y  =  24;  2- =  6. 

18    Given    \  £  +^Z^    to  find  »  and  y. 

Ans.  a?  =  2  ;  y  =  1. 

19.  Given    J  §+£  =  »»£    to  find  y  and  *. 

|C7*  In  this  example,  we  cannot  obtain  identical  terms  by  one 


$   128.]       EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  87 

multiplication.  But  we  may  apply  the  same  principle,  §105, 
that  is  used  for  finding  a  common  denominator.  For,  if  we 
multiply  the  co-efficient  of  the  first  equation  by  the  co-efficient 
of  the  second,  the  product  will  be  the  same  as  if  we  multiply 
the  co-efficient  of  the  second  by  the  co-efficient  of  the  first. 
Thus, 

Multiplying  the  first  by  4,  20?/  +  I2x  —  372 

Multiplying  the  second  by  3,  9y  +  12a?  =  240 

Subtracting  the  second  from  ~) 

i     c  i  r       liv  ==  i«* 

the  first,  and  unite,  3  u 

Ans.  y  =  12;  x  =  11. 
20.  Given    j  J^~a    ~7(    to  find  y  and  z. 

Multiplying  the  first  by  5,  ■   20y— 25z  =  10 

Multiplying  the  second  by  4,  20?/— 162:  =  28 

Subtracting  the  first  from  the  £ 
second,  and  unite,  3 

Ans.  ?/  =  3  ;  z  =  2. 

§  128.  From  the  foregoing,  we  derive  the  following: 

Rule  I.  to  exterminate  an  unknown  quantity. 

First,  Transpose,  so  as  to  bring  both  of  the  unknown 
quantities  to  the  left ;  x's  under  a?'s  ;  y's  under  ?/'s,  &c. 

Determine  which  of  the  unknown  quantities  you  will  ex- 
terminate; and  then,  if  it  is  necessary,  multiply  or  divide 
one  or  both  of  the  equations  so  as  to  make  the  term  which 
contains  that  unknown  quantify  to  be  the  same  in  both. 

Then  if  the  identical  terms  have  like  signs  in  both  equa- 
tions, subtract  one  equation  from  the  other;  but  if  they 
have  unlike  signs,  add  one  equation  to  the  other.  And  the 
result  will  be  an  equation  containing  only  one  unknown 
quantity. 

Questions.  What  is  the  first  operation  for  exterminating  an  un- 
known quantity  ]  Repeat  the  whole  rule.  What  are  identical  terms'? 
Why  do  we  add  them  when  the  signs  are  unlike  1  Why  do  we  sub- 
ract  them  when  the  signs  are  alike  ? 


ALGEBRA.  [SECTION  XI. 


EQUATIONS.— SECTION  11. 

1.  What  two  numbers  are  those  whose  sum  is  20  and  dif 
ference  12? 

Stating  the  question,  x  =  greater  number. 

y  =■  the  less. 

Then  forming  the  equations,  x+y  =  20 

x—y  =  12 

As  the  signs  of  the  identical  terms  ) 

?..        ....  .  >2x       =32.-.a?=16 

are  unlike,  ado  the  equations,     3 

Substituting  16  for  x  in  the  first,        16-f  y  =  20 

Transposing  and  uniting,  y  =  4. 

Ans.   16  and  4. 

2.  A  market  woman  sells  to.  one  person,  3  quinces  and  4 
melons  for  25  cents;  and  to  another,  4  quinces  and  2  melons, 
at  the  same  rate,  for  20  cents.  How  much  are  the  quinces 
and  melons  apiece  ? 

After  the  statement,  forming  the         3         3a?-j-4?/  =  25 
equations,  3  4a?-f2i/  =  20 

Multiplying  the  second  by  2,  8#-f  4y  =  40 

Subtracting  first  from  third,  5x  =15 

Ans.  Quinces,  3  cents  apiece ;  melons,  4. 
In  our  solutions  after  this,  we  shall  number  the  lines,  so 

that  any  reference  to  them  will  be  easily  understood. 

3.  A  man  bought  3  bushels  of  wheat  and  5  bushels  of  rye 
for  38  shillings ;  and  at  another  time,  6  bushels  of  wheat  and 
3  bushels  of  rye  for  48  shillings.  What  was  the  price  for  a 
bushel  of  each  ? 

Let  x  =  price  of  wheat,  and  y  =  price  of  rye. 

1.  By  the  first  condition,  3x  +  by  =  38 

2.  By  the  second,  6x  +  Sy  =  48 

3.  Multiply  the  1st  by  2,  6x+\0y  =76 

4.  Subtracting  the  2d  from  the  3d,"-       7y  =  28.-.y  —  4 

5.  Substituting  4  for  y  in  the  1st,     3x +20  =  38. 

Ans.  Wheat  for  6s. ;  rye  for  4* 


§  128.]       EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  89 

4.  Two  purses  together  contain  $400.  If  you  take  $40 
out  of  the  first  and  put  them  into  the  second,  then  there  is 
the  same  in  each.     How  many  dollars  does  each  contain  ? 

Let  x  =  the  number  in  the  first. 
y  =  the  number  in  the  second. 
gCT3*  Although  in  these  examples  we  have  omitted  the  state- 
ment, it  is  expected  the  pupil  will  state  them  as  usual. 

1.  By  the  first  condition,  x  +  y  =  400 

2.  By  the  second,  x — 40=y+40 

3.  Transposing  the  2d,  x  —  y  =   80 

4.  Adding  the  1st  to  the  3d,  ~2x  =480  .-.  x  =  240 

Ans.  The  first,  $240;  the  second,  $160. 

5.  A  gentleman  being  asked  the  age  of  his  two  sons,  re- 
plied, that  if  to  the  sum  of  their  ages  25  be  added,  this  sum 
will  be  double  the  age  of  the  eldest ;  but  if  8  be  taken  from 
the  difference  of  their  ages,  the  remainder  will  be  the  age  of 
the  youngest.     What  is  the  age  of  each  ? 

Let  x  =  the  age  of  the  eldest,     y  =  the  age  of  the  youngest. 

1.  By  the  first  condition,  x+y-\-25  =  2x 

2.  By  the  second,  x — y — 8=y 

3.  Transposing  and  uniting  1st,  — x-j-y  =  — 25 

4.  Transposing  and  uniting  2d,  x — 2y=        8 

5.  Adding  the  3d  and  4th,  —  y—— 17 

6.  Substituting  17  for  y  in  the  3d,  — a?-f 17  =  —25 

7.  Transposing,  — x  =  — 42 

Ans.  Eldest,  42 ;  youngest,  17. 

6.  A  gentleman  paid  for  6  pair  of  boots  and  4  pair  of  shoes 
$44 ;  and  afterwards  for  3  pair  of  boots  and  7  pair  of  shoes, 
$32.     What  was  the  price  of  each  per  pair  ? 

Ans.  Boots,  $6 ;  shoes,  $2. 

7.  A  man  spends  30  cents  for  apples  and  pears,  buying  his 
apples  at  the  rate  of  4  for  a  cent,  and  his4  pears  at  the  rate  of 
5  for  a  cent.  He  afterwards  let  his  friend  have  half  of  his 
apples  and  one-third  of  his  pears  for  13  cents,  at  the  same 
rate.     How  many  did  he  buy  of  each  sort  ? 

8* 


90 


ALGEBRA. 


[SECTION  XI. 


Let  x  =  number  of  apples. 
y  =  number  of  pears. 

-  cent  =  price  of  1  apple. 

-  cent  =  price  of  1  pear. 

x 

-  cents  =  price  of  all  the  apples 


.'/ 


1.  By  the  first  condition, 

2.  By  the  second, 

3.  Dividing  the  1st  by  3, 

4.  Subtracting  3d  from  2d, 

5.  Multiplying  by  24, 


-  cents  =  price  of  all  the  pears. 

x      y 
4+1=30 

8+15-13 
x       v 

12+T5=10 

X        X 

8~l2  =  3 
3x— 2x=72.-.a:=:72. 
Ans.  72  apples ;  60  pears. 


8.  One  day  a  gentleman  employs  4  men  and  8  boys  to 
labor  for  him,  and  pays  them  40s. ;  the  next  day  he  hires  at 
the  same  rate,  7  men  and  6  boys,  for  50s.  What  are  the  daily 
wages  of  each  ?  Ans.  Man's,  5s. ;  boy's,  2s.  6c?. 

9.  It  is  required  to  find  two  numbers  with  the  following 
properties  :  J-  of  the  first  with  -£  of  the  second  shall  make  16, 
and  ±  of  the  first  with  1  of  the  second  shall  make  9. 

$C7*  Performed  as  Problem  7.  Ans.  12  and  30 

10.  Says  A  to  B,  Give  me  5s.  of  your  money,  and  I  shall 
have  twice  as  much  as  you  will  have  left.  Says  B  to  A, 
Give  me  5s.  of  your  money,  and  I  shall  have  three  times  as 
much  as  you  will  have  left.     What  had  each  ? 

^7»  In  the  equations,  first  transpose  so  that  x  shall  be 
under  x,  and  y  under  y  Ans.  A,  lis.;  B,  13s 


§  128.]       EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  91 

11.  Two  men  agree  to  buy  a  house  for  $1200.  Says  A  to 
B,  Give  me  §  of  your  money,  and  I  shall  be  able  to  pay  for  it 
all ;  No,  says  B,  give  me  |  of  yours,  and  then  I  can  pay  for 
it.     How  much  money  had  each  ?     Ans.  A,  $800  ;  B,  $600. 

12.  Find  two  numbers  with  the  following  properties:  The 
products  of  the  first  by  2,  and  the  second  by  5,  when  added, 
are  equal  to  31 ;  also,  the  products  of  the  first  by  7,  and  the 
second  by  4,  when  added,  are  equal  to  68.         Ans.  8  and  3. 

13.  A  paid  B  20  guineas,  and  then  B  had  twice  as  much 
money  as  A  had  left ;  but  if  B  had  paid  A  20  guineas,  A  would 
have  had  three  times  as  much  as  B  had  left.  What  sum  did 
each  possess  at  first?  Ans.  A,  52  guineas ;  B,  44. 

14.  A  person  has  a  saddle  worth  £50,  and  two  horses. 
When  he  saddles  the  poorest  horse,  the  horse  and  saddle  are 
worth  twice  as  much  as  the  best  horse ;  but  when  he  saddles 
the  best,  he  with  the  saddle  is  worth  three  times  the  poorest. 
What  is  the  value  of  each  horse  ? 

Ans.  Best,  £40  ;  poorest,  £30. 

15.  A  merchant  sold  a  yard  of  broadcloth  and  3  yards  of 
velvet  for  $25 ;  and,  at  another  time,  4  yards  of  broadcloth 
and  5  yards  of  velvet  for  $65.  What  was  the  price  of  each 
per  yard?  Ans.  Broadcloth,  $10  ;  velvet,  $5. 

16.  A  person  has  500  coins,  consisting  of  eagles  and  dimes ; 
and  their  value  amounts  to  $1931.  How  many  has  he  of  each 
coin  ?         JC73*  The  solution  must  be  in  cents. 

Ans.  190  eagles;  310  dimes. 

17.  In  the  year  1299,  three  fat  oxen  and  six  sheep  together 
cost  79  shillings ;  and  the  price  of  an  ox  exceeded  the  price 
of  12  sheep  by  10  shillings.     What  was  the  value  of  each  ? 

Ans.  An  ox,  24s.;  a  sheep,  Is.  2d, 

.  18.  Two  persons  talking  of  their  ages,  A  says  to  B,  8  year* 
ago  I  was  three  times  as  old  as  you  were ;  and  4  years  hence, 
I  shall  be  only  twice  as  old  as  you  will  be.  What  are  their 
present  ages  ?  Ans.  A,  44  ;  B,  20  years. 


92  ALGEBRA.  [SECTION    XI. 

19.  A  farmer  sold  to  one  man  30  bushels  of  wheat  and  40 
of  barley  for  270  shillings ;  and  to  another,  50  bushels  of 
wheat  and  30  of  barley  for  340  shillings.  What  was  the  price 
per  bushel  of  each  ?  Ans.  Wheat,  5s. ;  barley,  3s. 

20.  A  man  and  his  wife  and  child  dine  together  at  an  inn. 
The  landlord  charged  15  cents  for  the  child,  and  for  the  woman 
he  charged  as  much  as  for  the  child  and  -}  as  much  as  for  the 
man ;  but  for  the  man  he  charged  as  much  as  for  the  woman 
and  child  together.     What  did  he  charge  for  each  ? 

Ans.  45  cents  for  the  man ;  and  30  cents  for  the  woman. 

21.  A  gentleman  has  two  horses,  and  also  a  chaise  worth 
$250.  If  the  first  horse  be  harnessed,  he  and  the  chaise  will 
be  worth  twice  as  much  as  the  second  horse ;  but  if  the  second 
be  harnessed,  he  and  the  chaise  will  be  worth  three  times  as 
much  as  the  first  horse.     What  is  the  value  of  each  horse  ? 

Ans.  First,  $150;  second,  $200. 

22.  A  is  in  debt  $1200,  and  B  owes  $2500;  but  neither 
has  enough  to  pay  his  debts.  A  says  to  B,  Lend  me  the  i 
of  your  fortune,  and  then  I  can  pay  my  debts.  But  B  an- 
swered, Lend  me  the  £  of  your  fortune,  and  I  can  pay  my 
debts.     What  was  the  fortune  of  each  ? 

Ans.  A,  $900;  B,  $2400. 

23.  A  wine  merchant  has  two  kinds  of  wine,  one  at  5s.  a 
gallon,  and  the  other  at  12s. ;  of  which  he  wishes  to  make  a 
mixture  of  20  gallons  that  shall  be  worth  8s.  a  gallon.  How 
many  gallons  of  each  sort  must  he  use  1 

Ans.  8f  gallons  of  that  at  12s. ;  1 1-f  of  that  at  5s. 


§  131.]  SECOND  METHOD  OF  EXTERMINATION.  93 

SECTION  XII. 
SECOND  METHOD  OF  EXTERMINATION. 

§  129.  In  each  of  the  preceding  questions,  we  first  found 
the  value  of  one  of  the  unknown  quantities  ;  and  then  substi- 
tuted that  value  for  that  unknown  quantity  in  one  of  the  equa 
tions,  in  order  to  find  the  value  of  the  other  unknown  quantity. 
This  mode  of  operating  furnishes  a  hint  that  leads  us  to  an- 
other method  of  extermination. 

Let  us  take  the  first  question  in  the  last  section,  [p.  88,] 

in  which  we  have  the  equations,    \  ~i9  £ 

The  last  part  of  our  operation  was  to  substitute  the  value 
of  x  for  x  itself,  in  one  of  the  equations.  It  is  evident  that 
we  could  make  this  substitution  just  as  well  if  the  value  of  x 
was  a  literal  quantity,  instead  of  16.     Thus,  supposing  x  to 

y 

be  equal  to  £  ;  then  substituting  it  for  x,  the  first  equation 

would  be  -^+i/=20. 

I 

§  130.  Let  us  therefore  transpose  the  first  equation  to  find 
what  x  will  equal,  just  as  if  we  knew  the  value  of  y.  We 
shall  find  that  x  =  20—  y.  And  then  in  the  second  equation, 
we  shall  use  the  value  of  x  instead  of  x  itself. 

Thus,  20— y— y=12. 
Transposing  and  uniting,  — 2y  =  — 8  .*.  y  =  4 ; 

which  was  our  answer  by  the  first  method.     Then  x  will  be 
found  by  substituting  4  for  y.     Whence  we  derive 

Rule  II.  to  exterminate  an  unknown  quantity. 

§  13 1.  Select  the  most  simple  term  of  the  unknown  quan- 
tities, and  by  the  equation  that  contains  it,  find  the  value 
of  that  unknown  quantity,  as  if  the  other  were  known;  and 


94  ALGEBRA.  [SECTION  XII. 

then,  in  the  other  equation,  substitute  this  value  for  the  un- 
known quantity  itself. 

We  shall  then  have  an  equation  with  only  one  unknown 
quantity  ;  which  may  be  solved  as  usual. 

EQUATIONS.— SECTION  12. 

1.  There  are  two  numbers  whose  sum  is  100 ;  and  three 
times  the  less  taken  from  twice  the  greater,  leaves  150  re- 
mainder.    What  are  those  numbers  ? 

Let   x  =  greater. 
y  =  less. 
2x — 3y  =  the  required  subtraction. 
Forming  the      f   1.  By  the  first  condition,  x+y=l00 
equations,      c  2.   By  the  second,  2x — 3*/=  150 

3.  Transposing  the  1st,  a?  =100— y 

4.  Multiplying  the  3d  by  2,  2a:  =  200— 2y 
6.  Substituting  200-2y  for  2*  >    800_8       3        160 

in  the  2d,  5  J       n 

6.  Transposing  and  uniting,        — 5t/= — 50.\i/=10 

7.  Substituting  10  for  y  in  the  1st,  x -j- 10  =  100 

8.  Transposing  and  uniting,  x  =  90 

Ans.  Greater,  90;  less,  10. 

2.  The  ages  of  a  father  and  his  son  amounted  to  140  years; 
and  the  age  of  the  father  was  to  the  age  of  the  son  as  3  to  2. 
What  were  their  ages  ? 

Let  x  =  age  of  the  father. 
y  =  age  of  the  son. 

1.  By  the  first  condition,  x+y-=.  140 

2x 

2.  By  the  second,  y  =  — 

o 

2x  2x 

3.  Substituting  —  for  y  in  the  1st,         a?+-— =  140 

3  3 

4.  Multiplying  by  3,  3#+2.r=420 

5.  Uniting  and  dividing,  x  =  84 
&  Substituting  84  for  x  in  1st,                 84-f-y  =  140 

Ans.  Father,  84  years  ;  son,  56 


§  131.]  SECOND   METHOD    OF    EXTERMINATION  95 

3.  Find  two  numbers,  such  that  -£  of  the  first  and  A  of  the 
second  shall  be  87 ;  and  \  of  the  first  and  \  of  the  second 
shall  be  55.  Ans.  135,  and- 168. 

4.  A  says  to  B,  Give  me  100  of  your  dollars,  and  I  shall 
have  as  much  as  you.  B  replies,  Give  me  100  of  your  dol- 
lars, and  I  shall  have  twice  as  much  as  you.  How  many 
dollars  has  each?  Ans.  A,  $500;  B,  $700. 

5.  There  are  two  numbers,  such  that  |  of  the  first  and  f 
of  the  second  added  together,  will  make  12;  and  if  the  first 
be  divided  by  2,  and  the  second  multiplied  by  3,  f  of  the  sum 
of  these  results  will  be  26.  Ans.   15,  and  10£. 

6.  Find  two  numbers  in  the  proportion  of  2  to  1,  so  that  if 
4  be  added  to  each,  their  two  sums  shall  be  in  proportion  of 
3  to  2.  Ans.  8,  and  4. 

7.  A  and  B  owned  9800  acres  of  western  land.  A  sells  J 
of  his,  and  B  sells  \  of  his  ;  and  they  then  have  just  as  much 
as  each  other.     How  many  acres  had  each  1 

Ans.  A,  4800;  B,  5000. 

8.  A  son  asking  his  father  how  old  he  was,  received  the 
following  reply :  My  age,  says  the  father,  7  years  ago,  was 
four  times  as  great  as  yours  at  that  time ;  but  7  years  hence, 
if  you  and  I  live,  my  age  will  be  only  double  of  yours. 
What  was  the  age  of  each  ? 

Ans.  Father's,  35  years;  son's,  14  years. 

8.  The  weight  of  the  head  of  Goliath's  spear  was  less  by 
one  pound  than  ■}-  the  weight  of  his  coat  of  mail ;  and  both 
together  weighed  17  pounds  less  than  ten  times  the  spear's 
head.     What  was  the  weight  of  each  ? 

Ans.  Coat,  208  pounds ;  spear's  head,  25  pounds. 

10.  A  market  woman  bought  eggs,  some  at  the  rate  of  2  for 
a  cent,  and  some  at  the  rate  of  3  for  2  cents,  to  the  amount  of 
65  cents.  She  afterwards  sold  them  all  for  120  cents,  thereby 
gaining  half  a  cent  on  each  egg.  How  many  of  each  kind  did 
she  buy  ?        Ans.  50  of  the  first  kind ;  60  of  the  other  kind 


96  ALGEBRA  [SECTION  XII. 

11.  Says  A  to  B,  |  of  the  difference  of  our  money  is  equal 
to  yours ;  and  if  you  give  me  $2,  I  shall  have  five  times  as 
much  as  you.    How  much  has  each  ?     Ans.  A,  $48 ;  B,  $12. 

12.  A  and  B  possess  together  property  to  the  amount  of 
$5700.  If  A's  property  were  worth  three  times  as  much  as 
it  is,  and  B's  five  times  as  much  as  it  is,  then  they  both  would 
be  worth  $23,500.     What  is  the  worth  of  each  ? 

Ans.  A,  $2500 ;  B,  $3200. 

13.  A  gentleman  has  two  silver  cups,  and  a  cover  adapted 
to  each  which  is  worth  $20.  If  the  cover  be  put  upon  the 
first  cup,  its  value  will  be  twice  that  of  the  second  ;  but  if  it 
be  put  upon  the  second,  its  value  will  be  three  times  that  of 
the  first.     What  is  the  value  of  each  cup  ? 

Ans.  First  cup,  $12;  second,  $16. 

14.  Two  men  driving  their  sheep  to  market,  A  says  to  B, 
Give  me  one  of  your  sheep,  and  I  shall  have  as  many  as  you. 
B  says  to  A,  Give  me  one  of  your  sheep,  and  I  shall  have 
twice  as  many  as  you.     How  many  had  each  ? 

Ans.  A,  5  sheep ;  B,  7. 

15.  What  two  numbers  are  those,  whose  difference  is  4, 
and  5  times  the  greater  is  to  6  times  the  less,  as  5  to  4  ? 

Ans.  8  and  12. 

16.  There  are  two  numbers  such  that  £  of  the  greater  added 
to  |  of  the  less,  will  equal  13  ;  and  if  i  of  the  less  be  taken 
from  i  of  the  greater,  the  remainder  is  nothing.  What  are 
the  numbers  ?  Ans.  18  and  12. 


§  133.]  THIRD    METHOD    OF    EXTERMINATION.  97 

SECTION  XIII. 

THIRD  METHOD  OF  EXTERMINATION. 

§  132.  The  method  of  substitution  as  explained  in  the  last 
section,  may  be  modified  a  little.  We  will  show  how,  by 
using  question  1st,  in  the  last  section  of  equations. 

lhe  two  equations  were    <  „   __q    1*0 

We  transpose  the  1st;  thus,  x  =  100 — y. 

Now,  before  we  substitute  the  value  of  x  for  x  itself  in  the 
second  equation,  we  will  transpose  the  second  equation  so  as 
to  make  x  stand  alone  ;  thus,         2x  —  1 50  +  3y. 

Then  substitute  the  value  of  x  as  found  before  by  the  first 
equation,  200— 2y  =  150-j-3t/  with  which  we  may 

proceed  as  before. 

§  133.  Before  we  make  the  substitution  after  transposing, 
it  is  generally  best  to  find  the  value  of  x  alone  in  the  second 
equation.     Thus, 

Given    \  C—2y  =  10  \    t0  find  X  and  y- 

Transposing  and  dividing  the  1st,     Xwm  ~    '  ■-&- 

2 

Transposing  and  dividing  the  2d,      x  =  — -— -  • 

Now,  as  it  is  evident  that  things  which  are  equal  to  the 
same,  are  equal  to  one  another ;  one  value  of  x  is  equal  to 
the  other  value  of  x ;  thus, 

23— 3y__  IO+2.7 
2    "  ~       5^" 
Destroying  the  fractions,  115 — 15^  =  20  +  42/ 

Transposing,  uniting,  and  dividing,  y  =  5 

By  substituting  the  value  of  y  in  one  of  the  equations,  we 
find  x  =  4.     Whence  we  derive 
G  9 


08  ALGEBRA.  [SECTION  XIII. 

Rule  III.  to  exterminate  an  unknown  quantity. 
§  134.  Find  by  each  of  the  equations,  the  value  of  that 
unknown  quantity  wlrich  is  the  least  involved;  and  then 
form  a  new  equatin  by  making  one  of  these  values  equal 
to  the  other. 

EQUATIONS.  — SECTION  13. 

1.  Divide  $60  between  A  and  B,  so  that  the  difference  be- 
tween A's  share  and  31,  may  be  to  the  difference  between  31 
and  B's  share,  as  6  to  7. 

Let  x  =  A's  share;  and?/=B's. 

1.  By  the  firs^  condition,  x+y  —  60 

2.  By  the  second,  x— 31  :  31—  y  ::.6  :  7 

3.  Multiplying  extremes  and  means,        7x — 217  m  186— 6y 

4.  Transposing  the  1st,  x  =  60— y 

5.  Transposing  and  uniting  the  3d,  7#  =  403— fiy 

6.  Multiplying  the  4th,  7a:  =  420— ly 

7.  Equating  5th  and  C.th,  403— 63/  =  420— ly 

8.  Transposing  and  uniting,  y  =  17 

0.  Substituting  17  in  the  4th,  x  =  60—17  =  43 

Ans.  A's  share,  $43;  B's,  $17. 

2.  There  is  a  fraction,  such  that  if  1  is  added  to  the  nume- 
rator, its  value  will  be  -£ ;  but  if  1  be  added  to  the  denomina- 
tor, its  value  will  be  \.     What  is  that  fraction  ? 

Let   x  =  numerator ;  and  y  =  denominator. 

x 
The  fraction  will  be, 

y 

x+\      1 

1.  By  the  first  condition,  =  — 

y       3 
x        1 

2.  Bv  the  second,  — rr—T 

y  +  l      4 

3.  Multiplying  the  1st  by  y%  and  by  3,         3#  +  3  —y 

4.  Multiplying  the  2d  by  y  +  l,  and  by  4,  ±x  =  y  +  \ 

5.  Transposing  the  4th,  4x — 1  =y 

6.  Equating  3d  and  5th,  3a?  +  3  =  4.r— 1 

7.  Transposing  and  uniting,  — x  = — 4 

8.  Substituting  the  value  of  3x  in  the  4th,         15  =y 

Ans.   £. 


§   134.]  THIRD    METHOD    OF    EXTERMINATION.  99 

3.  There  is  a  certain  number,  consisting  of  two  places  of 
figures,  which  is  equal  to  4  times  the  sum  of  its  digits ;  and 
if  18  be  added  to  it,  the  digits  will  be  inverted.  What  is  thaJ 
number  ? 

Let  x  =  first  digit  or  tens;  and  y  =  the  units 
10a? -fi/  =  the  number. 
4x-\-4y  =  four  times  the  sum  of  digits. 
1037+1/+ 18,  =  when  18  is  added. 

\0y-\-x,=  when  the  digits  are  inverted 

1.  By  the  first  condition,  \0x+y —  4x-\- 4y 

2.  By  the  second,  l0x-\-y+ 18  =  lOy+x 

3.  Transposing  and  uniting  the  1st,  6a?  =  dy 

4.  Transposing  and  uniting  the  2d,  9x  =  9y — 18 

5.  Multiplying  the  4th  by  f ,  6a?  =  6y— 12 

6.  Equating  3d  and  4th,  Sy  =  6y— 12 

Ans.  24. 

4.  There  is  a  certain  number  consisting  of  two  figures ;  and 
if  2  be  added  to  the  sum  of  its  digits,  the  amount  will  be  three 
times  the  first  digit;  and  if  18  be  added  to  the  number,  the 
digits  will  be  inverted.     What  is  the  number?  Ans.  46. 

5.  A  person  has  two  snuff-boxes  and  $8.  If  he  puts  the  8 
dollars  into  the  first,  then  it  is  half  as  valuable  as  the  other. 
But  if  he  puts  the  8  dollars  into  the  second,  then  the  second 
is  worth  three  times  as  much  as  the  first.  What  is  the  value 
of  each  ?  Ans.  First,  $24  ;  second,  $64. 

6.  A  gentleman  has  two  horses  and  a  chaise.  The  first 
horse  is  worth  $180.  If  the  first  horse  be  harnessed  to  the 
chaise,  they  will  together  be  worth  twice  as  much  as  the  se- 
cond horse ;  but  if  the  second  horse  be  harnessed,  the  horse 
and  chaise  will  be  worth  twice  and  one-half  the  value  of  the 
first.  What  is  the  value  of  the  second  horse,  and  of  the 
chaise ?  Ans.  Horse,  $210;  chaise,  $240. 

7.  There  is  a  certain  number  consisting  of  two  digits.  The 
sum  of  these  digits  is  5  ;  and  if  9  be  added  to  the  number  itself, 
the  digits  will  be  inverted.     What  is  the  number?     Ans.  23. 


100  ALGEBRA.  [SECTION  XIII. 

8.  There  is  a  number  consisting  of  two  figures.  If  the 
number  be  divided  by  the  sum  of  the  figures,  the  quotient  will 
be  4  ;  but  if  the  number  made  by  inverting  the  figures  be  di- 
vided by  1  more  than  their  sum,  the  quotient  will  be  6.  What 
is  the  number? 

|C?*  In  the  operation,  4  multiplied  by  x+y,  is  the  same  as 
4  times  x+y.  Ans.  24. 

9.  There  are  two  numbers  such  that  the  less  is  to  the 
greater  as  2  to  5  ;  and  the  product  made  by  multiplying  the 
two  numbers  together,  is  equal  to  ten  times  their  sum.  What 
are  the  numbers  ? 

Let  x  =  the  less  ;  and  y  =  the  greater. 

2y 

1.  By  the  first  condition,  x—* 

o 

Note. — If  we  wish  to  multiply  y  by  4,  we  put  4  immedi- 
ately before  the  y  as  a  co-efficient;  and  in  the  same  way,  if 
we  multiply  y  by  x,  we  make  x  the  co-efficient  of  y. 

2.  By  the  second,  xy=lf)x+l0y 

3.  Multiplying  the  1st  by  10,  I0x  =  4y 

4.  Transposing  the  2d,  I0x  =  xy — lOy 

5.  Equaling  4th  and  3d,  xy—l0y  =  4y 

Note. — When  we  divide  4x  by  4,  we  do  it  by  taking  away 
4  when  we  divide  10a;  by  10,  we  do  it  by  taking  away  the 
U  Tn  the  same  manner  we  divide  yx  by  y,  in  taking  away 
they 

6.  Divid.i a  by  y,  x— 10  =  4   .-.   37=14. 

7.  Substituui.e  14  for  x  in  the  3d,  140  =  4y  .-.  y  =  35. 

Ans.   14  and  35. 

10.  There  are  *wo  numbers,  whose  sum  is  the  |  part  of 
their  product;  and  ti.e  greater  is  to  the  less  as  3  to  2.  What 
are  those  numbers?  Ans.   15  and  10 


§135.]  EQUATIONS  WITH  SEVERAL  UNKNOWN'  QUANTITIES.    101 


SECTION   XIV. 

EQUATIONS  WITH  SEVERAL  UNKNOWN 
QUANTITIES. 

§  135.  When  there  are  three  or  more  unknown  quantities; 
first,  transpose  all  the  unknown  quantities  to  the  left,  and  write 
them  so  that  letters  of  the  same  kind  shall  he  under  each  other. 
Then,  combine  successively  one  of  the  equations  with  each  of 
the  others,  so  as  to  exterminate  'the  same  unknown  quantity 
from  each.  By  this  means  there  will  be  obtained  a  number 
of  equations  one  less  than  the  original  number.  With  these 
perform  the  same  process  as  before ;  and  proceed  in  this 
manner  till  there  is  but  one  equation  containing  only  one  un- 
known quantity;  which  may  be  solved  by  the  usual  rule. 
Then  by  substitution,  the  value  of  the  other  unknown  quan- 
tities may  be  found,  in  the  reverse  order  in  which  they  were 
exterminated.   . 

EXAMPLES. 

in-     a       ♦•     I  xtJft*  =  ?a  1 to  find  *«  y» 

1.  Given  the  equations  <  x-f-2?/  +  3z  =  16    \-  J 

jx  —  y—2z  =  —3j       an     / 

4.  Subtracting  1st  from  2d,  y-\-2z  =  7 

5.  Subtracting  3d  from  1st,        2y  +  3z  =  12 

6.  Multiplying  4th  by  2,  2i/  +  4z  =  14 

7.  Subtracting  5th  from  6th,  z  =  2 

8.  Substituting  value  of  z  in  4th,        y  =  3 

9.  Substituting  in  the  1st,  #  =  4 


Questions.  In  solving  equations  with  several  unknown  quantities, 
what  must  be  done  first]  Then  which  unknown  quantity  must  be 
exterminated  * 

9* 


/02  ALGEBRA.  [SECTION  XIV. 


rx  +  y  +  z  =  29  \ 
}ar+2y  +  3*  =  62/ 


x +  y  +  2  =  29 
8.  Given  }*+2y+3z  =  62  £   to  find  x, 

y,  and  2. 


4.  Subtracting  1st  from  2d,  1/4-22  =  33 

5.  Destroying  fractions  in  3d,  Qx  +  4y  +  3z  =  120 

6.  Multiplying  1st  by  6,  6x4-61/4-62  =  174 

7.  Subtracting  5th  from  6th,  2?/  +  3z  =  54~ 

8.  Multiplying  4th  by  2,  2y  +'4z  =  66 

9.  Subtracting  7th  from  8th,  z  =  12 
Whence  by  substitution,  y  =  9;  and  x  =  8. 

3.  Given   x+y+ ar  — 7j   2x  —  y—  32  =  3;   and  5x—3y 

4-52=19;  to  find  x,  y,  z.  Ans.  x  =  4,  i/  =  2,  2  =  1. 

4.  Given  a?— y— 2  =  5;  3x  +  4i/4-52  =  52 ;  and  5x—4y 
— 3z  =  32 ;  to  find  x,  y,  and  2.     Ans.  x  =  10,  y  =  3,  z  =  2. 

5.  Given   7x-f-5y+2z  =  79 ;    8x4-71/4-92=122;   and 
a?+4y -j-52  =  55  ;  to  find  the  values  of  x,  1/,  and  2. 

Ans.  x  =  4,  i/  =  9,  2  =  3. 

6.  Given  a: 4-y 4-2  =13;  x+y-f u  =  17;  x+2-f-tt  =  18; 
and  y+z+u  =  21  ;  to  find  the  values  of  x,  y,  2,  and  w. 

Ans.  x  =  2,  y=5,  2  =  6,  m=10. 


PART   II. 


LITERAL    ALGEBRA. 

GENERAL    PRINCIPLES. 

§  136.  The  algebraical  operations  which  we  have  hitherto 
treated  of,  belong  to  that  part  of  the  science  which  was  known 
to  the  ancients,  and  which  was  in  use  till  about  A.  D.  1600. 
About  that  time,  Franciscus  Vieta,  a  Frenchman,  introduced 
the  general  use  of  letters  into  Algebra,  (denoting  the  known 
quantities  in  a  problem  by  consonants,  and  the  unknown  ones 
by  vowels.) 

§  137.  This  improvement  gave  anew  aspect  to  the  science. 
So  that  now  algebra  is  rather  the  representation  of  arithmeti- 
cal results,  than  the  results  themselves.  And  therefore,'  its 
most  general  object  is  to  afford  means  for  investigating  the 
laws  of  calculation  for  every  description  of  numerical  questions. 

§  138.  Operations  with  numbers  cannot  furnish  general 
rules,  for  two  reasons.  In  the  first  place,  we  cannot,  by 
mere  inspection  of  the  results,  determine  how  they  were  ob- 
tained. Thus,  12  may  be  the  result,  either  of  multiplying  3 
by  4,  or  adding  5  to  7,  or  subtracting  8  from  20,  or  dividing 
48  by  4,  &c.  &c. 

And  in  the  second  place,  every  figure  in  an  arithmetical 
result  has  a  determinate  value  which  is  peculiar  to  itself;  and 
therefore  cannot  be  applied  to  any  other  question. 

Questions.  How  does  modern  algebra  differ  from  the  ancient 
methods  1  Who  introduced  the  modern  method  1  What  is  the 
modern  use  of  algebra]  What  are  the  two  reasons  why  we  cannot 
obtain  general  rules  by  arithmetical  operations  1 

103 


104  ALGEBRA. 

§  139.  But  in  algebra,  the  letters  which  represent  thfl 
quantities,  retain  their  identity  throughout  the  whole  reason- 
ing ;  and  as  the  operations!  of  addition,  subtraction,  multioli- 
cation,  &c,  are  only  represented  by  signs,  we  readily  see  the 
depend-  nee  which  the  several  quantities  have  upon  one  an- 
other  And  as  the  result  is  represented  by  letters,  each  of 
w'  .ch  may  stand  for  whatever  number  we  choose,  its  value 
.8  entirely  indeterminate;  and  shows  merely  what  operations 
it  is  necessary  to  perform  upon  the  numbers  when  a  particular 
value  may  be  assigned  to  them. 

§  140.  Hence,  in  Literal  Algebra,  the  result  does  not  de- 
pend upon  the  particular  values  of  the  quantities  which  we 
operate  with,  but  rather  upon  the  nature  of  /he  question; 
and  it  will  always  be  the  same  for  every  question  of  the  same 
kind.     The  result  is  therefore  a  general  rule. 

§  141.  The  principal  signs  that  are  used  in  algebra  are  the 
following. 

The  sign  -f  {plus)  represents  addition.  Thus,  a-f  b  de- 
notes that  b  is  added  to  a.     a-{-b  is  the  sum  of  two  numbers. 

The  sign  —  (minus)  denotes  that  the  quantity  following  it 
is  subtracted.  Thus,  a—b  is  the  remainder  obtained  by  sub- 
tracting b  from  a. 

The  sign  x  (multiplied  by)  denotes  that  the  quantity  be- 
fore it,  is  multiplied  by  the  quantity  that  follows  it.  Thus, 
axb  is  the  product  of  a  and  b. 

The  sign  .  is  sometimes  put  between  two  literal  quantities 
instead  of  X  ;  as  a,b.  But  more  generally,  in  algebra,  the 
letters  are  joined  together,  to  represent  their  product.  Thus, 
ab  is  the  product  of  two  numbers. 

The  sign  -r-  or  :  (divided  by)  represent  the  division  of  the 

Questions.  How  does  algebra  differ  in  these  particulars?  In  lite- 
ral algebra,  what  is  the  use  of  the  answers?  Upon  what  does  an 
algebraical  result  depend  ?  In  what  cases  are  the  results  the  same  ? 
Define  the  signs  -j-,  — ,  X,  •,  -4-.  What  is  the  sum  of  two  num- 
bers, say  «  and  b  ?  What  is  their  difference  ?  What  is  their  pro- 
iuct  ?     What  is  their  quotient  ? 


$141.]  GENERAL    PRINCIPLES.  105 

quantity  before  it,  by  the  quantity  which  follows  it.  Thus, 
a—b  is  a  divided  by  b. 

Division  is  more  generally  denoted  in  algebra  by  writing 
the  divisor  under  the  dividend,  in  the  form  of  a  fraction. 

Thus,  7-  is  the  quotient  of  a  divided  by  b. 

The  sign  =  (equals)  denotes  that  the  whole  quantity  on 
the  left  of  it,  is  equal  to  the  whole  quantity  on  the  right  of  it. 
Thus,  4  +  8  =  16—4. 

The  sign  ±  or  =p  {plus  or  minus  and  minus  or  plus) 
shows  that  either  by  addition  or  by  subtraction,  the  effect  will 
be  the  same.  Thus,  there  are  circumstances  when  x  =  ±  a ; 
that  is,  x  is  equal  to  plus  a  or  minus  a. 

The  sign  ^  or  </d  (the  difference  of)  shows  that  it  is  not 
known  whether  the  quantity  before  it  is  subtracted  from  the 
quantity  after  it,  or  the  latter  is  subtracted  from  the  former. 
Thus,  a  \n  b  is  the  difference  between  a  and  6,  without  speci- 
fying which  is  the  greatest. 

The  sign  >  or  <  (greater  than  or  less  than)  denotes  that 
the  quantity  towards  which  it  opens  is  greater  than  the  other. 
Thus,  in  a  >  6,  a  is  greater  than  b. 

The  sign  (a  vinculum)  denotes  that  all  which  is 

put  under  it,  is  to  be  used  as  one  term.  Thus,  a-j-6 — ex 3, 
signifies  that  the  whole  quantity  under  the  vinculum,  is  to  be 
multiplied  by  3. 

The  line  which  separates  the  terms  of  a  fraction  is  also  a 

vinculum.     Thus,  — signifies  that  the  whole  quantity 

above  the  line  is  to  be  divided  by  the  whole  quantity  under  it. 
The  (  )  parenthesis  is  frequently  used  instead  of  the  vin 
culum ;  thus,  (a+b).  (c-j-e?)  signifies  the  product  of  (a-f-6) 
multiplied  by  (c  +  d). 


Questions.     Define  the  sign   =,  ±»   ^f,  <*> ,  >  <>  » 

(  ).     Is  the  line  above  a  quantity  the  only  vinculum!     What  is  tho 
sign  of  multiplication  between  vinculums  1 


106  ALGEBRA. 

The  sign  oo  {infinity)  denotes  a  quantity  that  is  infinitely 
large,  or  a  quantity  so  great  that  it  may  be  considered  larger 
than  any  supposable  quantity. 

The  cipher  0  is  sometimes  used  to  represent  a  quantity 
that  is  less  than  any  quantity  that  may  be  mentioned.  This 
is  always  the  case  when  the  cipher  is  used  as  a  denominator 
of  a  fraction. 

The  radical  sign  </  denotes  the  root  of  the  following  quan- 
tity. When  unaccompanied  by  a  figure,  it  represents  the 
square  root.  But  when  a  figure  is  put  over  it,  that  figure  ex- 
presses the  root  that  is  designed.  Thus,  $/a  denotes  the  3d 
root  of  a.  The  number  over  the  radical  sign  is  called  the 
index  of  the  root. 

A  co-efficient  is  a  number  put  immediately  before  a  letter, 
as,  2b ;  and  denotes  how  many  times  the  quantity  is  to  be 
taken. 

In  such  a  case,  the  co-efficient  is  called  a  numeral  co-effi-' 
cient.  Sometimes  when  one  letter  has  been  multiplied  into 
another,  the  first  written  one  is  called  a  literal  co-efficient; 
as,  in  ab,  a  is  the  literal  co-efficient. 

An  exponent  or  index  is  a  small  figure  placed  a  little  over 
and  a  little  to  the  right  of  a  quantity;  as,  a2.  It  denotes  that 
the  quantity  is  multiplied  by  itself;  and  shows  how  many 
times  the  quantity  is  to  be  taken  as  a  factor. 

An  exponent  may  be  either  positive,  negative,  or  fractional; 
as,  a2,  a-2,  a*.     This  will  be  explained  in  another  place. 

The  sign  : :  represents  proportion,  and  :  denotes  the  ratio 
of  two  numbers. 

The  sign  oc  denotes  a  general  proportion. 

A  term  is  any  quantity  that  is  not  separated  into  parts  by 
either  of  the  signs. 

Questions.  Define  the  signs  oo,  0,  >/,  <§/ >  What  is  the  figure 
over  the  radical  sign  called  ?  What  is  a  co-efficient?  Is  it  numeral 
or  literal?  What  is  an  exponent  or  index?  What  is  the  sign 
for  proportion  ?  and  for  ratio  ?  What  is  the  sign  for  a  general  pro- 
portion?    What  is  a  term  ? 


§   141.]  GENERAL    PRINCIPLES.  107 

The  sign  .*.  is  used  tor  the  word  therefore 

The  hyphen  -  is  used  for  the  word  which;  as,  2#  =  ^4 
-  =  8  ;  read,  ivhich  equals  8. 

A  simple  quantity  is  that  which  is  represented  by  one 
term ;  as,  o,  26,  — 3,  Smnrs,  &c. 

A  compound  quantity  is  that  which  is  represented  by  two 
or  more  terms;  as,  a-\-2ab—x. 

Similar  quantities  are  those  which  consist  of  the  same 
letters,  or  combinations  of  letters ;  as,  a  and  2« ;  Qbx  and  4bx. 

Dissimilar  quantities  are  those  which  consist  of  different 
letters,  or  different  combinations  of  them ;  as,  a  and  a2;  2ax 
and  2ab. 

Identical  terms  are  those  which  are  not  only  similar,  but 
also  have  the  same  co-efficient ;  as,  2ax  and  2ax. 

Identical  expressions  are  sometimes  made  up  of  the  same 
letters,  but  differently  combined  in  their  terms ;  as,  (3a — 26) 
and  (2a+a+ 56— 76). 

Positive  quantities  are  those  which  have  the  sign  -f-  before 
them,  either  expressed  or  understood  ;  as,  ab,   -\-ax. 

Negative  quantities  are  those  which  have  the  sign  —  be- 
fore them  ;  as*,  — 3,  —2x. 

Given  quantities  are  such  as  have  known  values ;  and  are 
generally  represented  by  the  first  letters  of  the  alphabet. 
They  are  sometimes  represented  by  the  initial  of  the  names 
that  stand  for  them  ;  as  s  for  the  sum;  d  for  the  difference. 

Unknown  quantities  are  those  which  are  to  be  discovered ; 
and  are  generally  represented  by  some  of  the  final  letters  of 
the  alphabet. 

A  quantity,  when  represented  by  one  term,  is  sometimes 
called  a  nomial.     When  it  has  two  terms,  it  is  called  a  bi- 


Quesiions.  What  is  the  sign  for  therefore  ]  For  which  ?  What 
is  a  simple  quantity]  What  is  a  compound  quantity?  'What  are 
similar  quantities'?  Dissimilar  quantities  ]  Identical  terms ?  Posi- 
tive quantities'?  Negative  quantities'?  Given  quantities]  Un- 
known quantities  ?     What  is  a  nomial  ]     Binomial  1 


108  ALGEBRA.  [EQ.  SEC.  1  &  2. 

nomial;  when  it  has  three,  it  is  called  a  trinomial  $  when 
it  has  many,  it  is  called  a  multinomial  or  polynomial. 

Each  of  the  literal  factors  which  compose  a  term,  is  called 
a  dimension  of  that  term  ;  and  the  number  of  these  dimensions 
or  factors  is  called  the  degree  of  the -term.  The  co-efficient 
is  not  counted  as  a  dimension.  Thus,  2a  is  a  term  of  one 
dimension  or  of  the  first  degree  ;  6ax  is  a  term  of  two  dimen- 
sions or  of  the  second  degree  ;  5a2; r2  is  a  term  of  four  dimen- 
sions or  of  the  fourth  degree.  And,  generally,  the  degree  of 
the  term  is  the  sum  of  the  exponents  which  belong  to  its 
letters. 

A  polynomial  is  called  homogeneous  when  all  its  terms 
are  of  the  same  degree.  Thus,  3abc  — ax2-f-c3,  is  homoge- 
neous ;  but  8a3 — 4ab+c  is  not  homogeneous. 

§  142.  The  use  of  the  foregoing  signs  makes  algebra  a 
species  of  language  which  brings  our  reasonings  into  a  very 
small  space ;  so  that  in  solving  a  problem,  or  demonstrating 
the  existence  of  a  numerical  relation,  the  connection  of  the 
several  ideas  is  perceived  with  great  facility. 

Example  \.  The  sum  of  $660  was  subscribed  for  a  certain 
purpose,  by  two  persons,  A  and  B ;  of  which  B  gave  twice 
as  much  as  A.     What  did  each  of  them  subscribe  ? 

Now,  a  question  similar  to  this,  and  with  the  same  numbers, 
was  solved  in  the  First  section  of  Equations,  on  page  25. 
We  will  solve  this  in  the  same  manner,  with  the  exception 
of  using  a  instead  of  660. 

Stating  the  question,  x  a  what  A  gave. 

2x  =  what  B  gave. 

Both  together  gave  x-\-2x;  also  they  gave  a  dollars. 
Forming  the  equation,  x+2x  =  a 

Uniting  the  terms,  Sx  =  a 

Dividing  by  3,  x  =  - 

3 

Questions.  What  is  a  trinomial  ?  Polynomial  1  What  is  the 
dimension  of  a  term  1  What  is  the  degree  of  a  term  %  When  is  a 
quantity  homogeneous?     What  is  the  use  of  algebraic  signs  1 


§   142.]  GENERAL    PRINCIPLES.  109 

Here  we  find  that  A  subscribed  ±  of  a,  which  at  this  time 
stands  for  $660. 

But  it  is  very  plain  that  we  would  solve  the  question  in  the 
very  same  manner,  if  the  sum  were  $240.  And  in  that  case 
a  would  stand  for  $240 ;  and  A's  share  of  it  would  be  $80, 
and  B's  share,  $160. 

In  the  same  manner,  if  the  sum  were  $360 ;  then  a  would 
stand  for  $360,  and  A's  share  would  be  }  of  $360;  that 
is,  $120.  And  in  the  same  manner  we  may  make  a  repre- 
sent any  sum;  and  still  A's  share  of  it  would  be  i  of  it. 
Hence,  this  substitution  of  a  letter  for  a  number,  is  called 
generalizing  the  operation. 

We  see  that  this  result  has  given  us  a  general  ride  for  di- 
viding any  sum  between  two,  so  that  one  of  them  shall  have 
twice  as  much  as  the  other.  The  rule  is,  The  least  share 
shall  be  one-third  of  the  sum  ;  and  the  greatest  share,  two- 
thirds  of  it.  ^ 

Problems. — In  the  same  manner  generalize  all  the  pro- 
blems in  the  First  section  of  Equations }  page  25. 

Example  2.  What  number  is  that,  which,  with  5  added  to 
it,  will  be  equal  to  40  ? 

This  is  the  first  problem  in  section  2,  which  we  will  gene- 
ralize ;  using  a  for  40,  and  b  for  5. 

Stating  the  question,  x  =  the  number. 

x-\-b  =  after  adding. 
Forming  the  equation,  a?-f  b  =  a 

Transposing  /»,  x  =  a — b. 

We  see  that  the  answer  is  found  by  subtracting  the  5  from 
the  40.     Thus,  40—5  =  35. 

Example  3.  Generalize  problem  3  of  section  2.  It  will 
be  found  that  the  literal  answer  is  the  same  as  in  example  2 ; 
because  the  questions  are  similar.  In  this  example,  a  repre- 
sents 23,  and  /;  represents  9.     Whence  x  =  23 — 9  -  =  14. 


Question.    What  do  we  call  the  generalization  of  an  operation  1 
10 


x+x+b= 

both  shares. 

x+x+b = 

a 

x+x  = 

a-b 

2x  = 

a—b 

x  — 

a — b 
~~2"-' 

a 
or,-- 

h 
_2 

110  ALGEBRA.  [EQ.  SEC.  1.  &  2 

Example  4.    Divide  17  dollars  between  two  persons,  so 
that  one  may  have  $4  more  than  the  other.  [Prob.  4,  sec.  2.] 
Let  IT  be  represented  by  a,  and  4  by  b. 

Stating  the  question,  ,    I  x  =  the  least  share. 

x+ b  =  the  greater. 

Forming  the  equation, 
Transposing  6, 
Uniting  terms, 

Dividing  by  2 

The  answer  is  found  by  subtracting  the  difference  or  4, 
from  the  whole  sum,  and  then  dividing  by  2. 

And  this  is  the  rule  for  all  similar  sums. 

The  5th  question,  on  page  32,  is  similar  to  the  one  just 
performed  ;  and  in  that,  a  represents  55,  and  /;  represents  the 
difference  or  7.  The  numerical  answer  is  found  by  the  rule 
just  shown. 

^        a—b      55—7         48 
Thus,  _,_.»_.  =24. 

§  143.  As  this  rule  is  of  some  importance,  it  will  be  well 
to  remember  it.  If,  from  a  number  to  be  divided  into  two 
parts,  we  subtract  the  difference  of  those  parts,  half  the  re- 
mainder will  be  equal  to  the  smaller  part. 

Applicati  m.    Perform  the  7th  and  10th  by  this  rule. 

Example  5.    In  the  same  questions,  let  us  take  x  for  the 
greatest  share.     Then  x — b  =  the  less. 
Forming  the  equation,  a?-f  x — b  =  a 

Transposing  and  uniting,  2x  =  a+b 

Dividing  by  2,  *  =  aJr,°r'2i  +  2:- 

§  144.  Here  we  have  another  rule.  If,  to  a  number  to 
be  divided  into  two  parts,  we  add  the  difference  between 
those  parts,  half  the  sum  will  be  equal  to  the  greater  part 


§  145.]  GENERAL    PRINCIPLES.  Ill 

Application.  Find  the  greater  part  in  questions  4,  5,  7, 
and  10,  on  page  32,  by  this  rule ;  without  algebra. 

§  145.  The  mere  letters  in  the  answer  of  an  algebraical 
operation,  form  what  is  called  a  formula ;  because  they  are 
the  form  of  the  solutions  of  all  similar  questions.  And  this 
is  the  advantage  of  representing  the  quantities  by  letters. 
For,  as  arithmetical  operations  on  them  can  only  be  indicated, 
the  result  also  must  be  merely  an  indication ;  and  this  indica- 
tion will  apply  to  any  question,  in  the  enunciation  of  which 
the  only  things  which  vary,  are  the  numerical  values  of  the 

s-i-d 
quantities.      Thus,   the  formula  x  = ,  denotes   that  the 

z 

greater  share  is  found  by  adding  the  difference  to  the  sum, 

and  dividing  the  amount  by  2. 

Example  6.    The  learner  must  now  generalize  problem  6, 

on  page  32  ;  using  b  and  c  for  the  two  differences.     The 

formula  that  he  obtains  will  be  the  answer  for  questions  8,  9, 

1 1,  and  13.    And  in  each  of  the  five  problems  the  pupil  must 

verify  the  answer  by  substituting  the  given  quantities  for  the 

i  ♦♦  m,  a— 2b— c        73-8—5 

letters.     Thus,  x  = = =  20. 

3  u 

We  have  said  that  in  algebra  the  arithmetical  operations  on 
numbers  are  only  represented  by  different  methods  of  com- 
bining the  signs  that  stand  for  those  quantities.  And  now, 
although  we  have  shown  in  our  progress  thus  far,  what  some 
of  those  methods  are,  it  may  be  well  to  review  them  a  little. 


Questions.  What  two  important  rules  have  we  found  by  gene- 
ralizing1? WThat  do  we  call  a  formula  in  algebra?  Why]  What 
advantage  do  we  derive  from  formulas?  What  then  is  the  real 
effect  of  an  algebraical  result  ? 


112  ALGEBRA.  [EQ.  SEC.  3  &  4. 


I. 

ADDITION  AND  SUBTRACTION  OF  ALGEBRAICAL 
QUANTITIES. 

§  146.  One  algebraical  quantity  is  added  to  another  by 
writing  one  quantity  after  the  other,  taking  care  to  preserve 
to  each  term  its  respective  sign.  Thus,  a+f—  c  is  added  to 
d—e+b,  so  as  to  make  d—  e-f  6-f-a+/—  c.  Or,  as  it  is 
easier  to  read  the  letters  in  their  alphabetical  order,  their  sum 
may  be  written  a  +  b — c  +  d— e-f/.  Again,  when  —a  is 
added  to  b,  we  preserve  the  sign ;  thus,  b — a. 

§  147.  One  algebraical  quantity  is  subtracted  from,  an- 
other, oy  changing  the  sign  or  signs  of  the  quantity  which 
is  to  be  subtracted,  and  then  writing  that  quantify  after  the 
other.  Thus,  a  +  h — y  is  subtracted  from  b — x-\-c,  by  first 
making  it  —a — h+y,  and  then  writing  the  whole  quantity, 
b— x+c— a— h+y;  or,  b— a+c— h-\-y— x. 

§  148.  After  the  addition  or  subtraction  has  been  performed, 
if  there  are  any  similar  quantities  in  the  result,  they  may  be 
united  by  adding  the  co-efficients  f  all  the  positive  similar 
terms,  and  affixing  their  literal  part  j  then  adding  all  the 
negative  similar  terms  in  the  same  manner ;  and  then  sub- 
tracting the  less  sum  from  the  greater,  and  retaini  >g,  in 
the  result,  the  sign  of  the  greater. 

§  149.  In  uniting  the  terms  of  compound  numbers,  we 
consider  the  literal  part  of  the  term  as  a  unit;  thus,  2a "and 
3a,  are  regarded  as  2  nnits  and  3  units  of  a  particular  kind 
which  when  put  together,  make  five  units  of  that  kind.  Now 
we  have  seen,  §  141,  that  the  co-efficient  of  a  quantity  may 

Questions.  How  is  addition  performed  in  algebra1?  How  is  a 
negative  quantity  addpd  1  How  is  subtraction  perform?  d  ?  How 
are  algebraical  quantities  united  1  In  determining  whether  the 
quantities  are  similar,  what  part  of  the  quantity  do  we  examine7 


§152.]  ADDITION    AND    SUBTRACTION.  113 

also  be  literal ;  as  in  ba,  ca,  &c.  In  such  cases,  the  whole 
term  ba  or  ca  becomes  a  unit,  each  of  a  different  kind ;  and 
of  course  are  not  similar  quantities,  and  cannot  be  united. 

§  150.  But  if  there  are  several  similar  units  of  this  kind, 
they  may  be  united  by  the  general  rule.  Thus,  ba — ca-\-ba 
ca-\-ba-\-ca,  can  be  united  into,  3ba  +  ca.  ax — bx -f ax -f  2bx 
— 3ax+bx,  are  equal  to, — ax-\-2bx;  or  2bx^-ax. 

§151.  Again,  we  have  seen,  §  68,  that  several  quantities 
are  sometimes  united  by  a  vinculum.  In  such  cases,  all  that 
is  embraced  by  the  vinculum,  is  regarded  as  a  unit  of  that 
kind ;  and  may  have  a  co-efficient.  Thus,  in  the  expressions, 
3X0— b+x,  and  5(x+ax— y);  a—b-\-x  is  a  quantity  taken 
3  times,  and  x+ax—y  is  a  quantity  taken  5  times.  Like 
quantities  of  this  kind  can  be  united;  thus,  2[ay— bx+x) 
-\-D(ay—bx+x)—7(ay—bx+x). 

§  152.  In  uniting  terms,  great  care  must  be  taken  that  the 
literal  part  be  entirely  alike,  both  in  signs  and  letters.  Thus, 
2bx-\-3cx,  cannot  be  united.  Neither  can  3y—2ay;  nor, 
6(a+bx)+2(ax+bx);  nor,  3.  ay— by  -f  2.ay2—by ;  nor, 
4{ax — bx)  —  2(ax~\- bx) ;  neither  in  any  other  case  where 
there  is  the  least  difference  in  any  part  but  the  leading  co- 
efficient. 

EXAMPLES. 

Unite  the  following  quantities. 

1.  3ax— 2y+4ax— 5?/-f ax— 3y.  Ans.  Sax—lOy. 

2.  3x-\-ay — 2a? — ay+4x  +  3ay  —  2x+4ay. 

Ans.  3x+lay. 

3.  4ax—y+3ay—2  —  2ax-\-ay—7y+8+2ay+y. 

Ans.  2ax-\-  Gay— 7y-f6. 

4.  ax — ay* — 3ay-\-  Sax — 2ay+7ay — 4ax — Say2. 

Ans.  2ax+2ay  —  9ay3. 


Questions.     Give  an  example  in  which  literal  quantities  are  not 
similar.     What  is  said  of  quantities  in  a  vinculum]     In  uniting, 
what  particular  care  is  necessary  1 
H  l0* 


114  ALGEBRA.  [EQ.  SEC.  3  &  4 

5.  8(a-y)+4(fl-y)+2(a-y)+7(a-y). 

Ans.  16(a— y) 

6.  —  4(a+6)+3(a+6)  —  2(a+6)+7(a+6).  


Ans.  4.  a-f  6. 

7.  2(a6+a?)-f3(aa?+6)  —  4(a?— #)  — 2(ab+x). 

Ans.  3(aa?+6) — 4(a?«— y). 

8.  7y-  4(a  +  b)  +  6y  +  2y  +  2(a  +  b)  +  (a  +  5)  +  y 
-3(a  +  5).  Ans.  16y— 4(a+6). 

9.  a?3  +  aa?-—  ab+ab —  a?3-f  a?y-faa?  +  a?y —  4ab  +  xa+x* 
—  x+xy+xy+ax.  Ans.  2x*+3ax  —  4ab+4xy  —  x. 

§  153.  Sometimes  the  subtraction  is  expressed  by  enclosing 
the  quantity  to  be  subtracted  in  a  parenthesis,  and  prefixing 
the  sign  — .  Thus,  4a  —  2x+3ax  —  (4x-\-3ay  —  2ax).  In 
such  cases,  when  the  vinculum  is  destroyed,  the  signs  must 
be  changed.  Thus,  4a  —  2x+3ax  —  4a?  —  3ay-\-2ax.  But 
if  there  is  a  co-efficient  immediately  before  the  vinculum,  the 
vinculum  cannot  be  destroyed,  nor  the  signs  changed ;  as  in 
ax  —  2(ab  —  3a?).  Because  such  quantities  are  considered  as 
only  one  unit.    §  151. 

10.  #+12  —  ax+y  —  (48  —  x  —  ax+3y). 

IC^First  subtract.     See  page  68.        Ans.  2a?  — 36  — 2y. 

11.  ab  —  4xy  —  a—  a?3  —  {2xy  —  6  + 14a? -fa:8). 

Ans.  a6  — 6a?y  —  a +6  —  14a?  —  2a?9. 

12.  3(a?+y)  +  (4.a7^/).      _  Ans.  7(*?+y). 


13.  2(a+b)  —  x  —  (3.a+6-a?3).     [See  §153.] 

Ans.  a?3  —  (a-j-b)  —  a?. 


14.  From  4.a+6,  take  a-f& —  3.x  — y.  

Ans.  3.a-f6  +  3.a?  — y. 

15.  a+b  —  (2a  — 3b)  -  (5a+7&)  —  ( —  13a-f2&). 

Ans.  la  —  56. 

Questions.  How  may  subtraction  be  expressed?  What  if  the 
vinculum  in  such  expressions  is  destroyed  1  When  cannot  the  vin- 
culum be  destroved  1     Why  1 


§  155.]  ADDITION    AND    SUBTRACTION  115 

16.  37a  —  5x  —  (3a  —  2b  —  5c)  —  (6a— 4&  +  3A). 

Ans.  28a-J-6&  —  5x-f5c  —  Sh. 

§  1 54.  After  the  subtraction  of  a  quantity  has  been  per- 
formed, we  may  transform  the  expression  by  changing  the 
signs  to  their  original  form  and  resupplying  the  parenthesis. 
Thus,  2x —  (3a-f2y — x),   becomes,  when   subtracted,  2x 

—  3a  —  2y+x;  and  this  latter  expression  may  be  restored 
back  to  2a?  —  (3a-\-2y  —  x). 

By  the  same  principle,  in  any  quantity,  we  may  suppose 
there  has  been  a  subtraction,  and  therefore  transform  the  ex- 
pression to  what  it  may  have  been. 

Thus,     2x— 3y+la+ax—  1, 
may  be  changed  to  2x — (3y — la — ax+\). 

ab  —  3x  —4y  — 2ax  -f  3a, 
may  become  either  ab  —  3x—4y — (2ax — 3a) 

or  ab  —  3x— (4y-j-2ax — 3a) 

or  ab  —  (3x+4y+2ax—  3a). 

§  155-  When  similar  quantities  have  literal  co-efficients; 
as,  mx  +  nx,  ay2 — by2,  <fcc. ;  a  compound  quantity  may  be 
expressed  by  placing  the  co-efficients  of  the  similar  quantities 
one  after  another,  (with  their  proper  signs,)  in  a  parenthesis ; 
and  then  annexing  their  common  letter  or  letters.  Thus, 
mx+nx  maybe  expressed  (m-\- n)x;  ay*— by*  by  (a — b)y2; 
axy-\-byx — abxy  by  (a -\-b — ab)xy,  &c. 

17.  From  ax  —  rx  +  ny  —  y8,  take  (a  —  n)y  -f  (a  -f  r)x 

—  (l-f-a)i/3.  Ans.  2ny  —  2rx  —  ay-\-ay3. 

18.  From  a(x-\-y)  -f  b (x  -fy),  take  c(x  -f  y)  —  d(x  —  y) 
+x  —  y.  Ans.  (a+b  —  c).(x+y)  +  (d—  l).(x  —  y). 

Questions.  When  a  quantity  has  been  subtracted,  how  can  it  be 
added  again  so  as  to  preserve  all  the  terms  ?  Give  an  example. 
How  will  this  principle  enable  us  to  transform  an  algebraical  ex- 
pression 1  Give  an  example.  What  may  be  done  with  literal  co- 
efficients i 


116  ALGEBRA.  [EQ.  SEC.  3  &  4. 

§156.  By  §154,  we  determined  that  1 — rx — tx  is  the 
same  as  1  —  (rx-\-tx),  which  by  §155  is  1 — (r-j-£)x;  and 
universally,  when  we  wish  to  enclose  literal  co-efficients  in  a 
parenthesis ;  if  the  first  of  them  has  the  sign  — ,  that  sign 
may  be  put  before  the  parenthesis,  and  the  signs  of  all  the 
enclosed  quantities  changed.  Thus,  1 — mx-\-px — qx+arx 
may  be  written  1 — (m — p-\- q — ar)x. 

§  157.  By  §  146,  — b  is  added  to  a,  so  as  to  make  a — b. 
Whence  it  will  be  seen  that  an  algebraical  sum  is  different 
from  an  arithmetical  sum ;  and  signifies,  not  that  it  is  greater 
than  one  of  its  parts,  but  that  it  is  the  result  of  putting  to- 
gether two  or  more  different  values,  either  of  the  same  or  of  a 
contrary  signification.  The  same  method  is  employed  in 
common  language,  when,  in  order  to  show  what  a  man's  for- 
tune is,  we  declare  what  is  owed  to  him  and  what  he  owes. 
In  both  an  algebraical  sum,  and  an  inventory  of  a  man's 
effects,  the  real  positive  value  may  be  much  smaller  than  any 
one  item  in  the  account,  and  even  less  than  nothing. 

§158.  Hence,  it  may  be  that,  after  uniting  terms,  we 
shall  have  a  negative  quantity,  — a.  This  shows  that,  having 
a  quantity  to  be  subtracted,  we  did  subtract  all  that  we  could, 
and  have  yet  more  to  be  subtracted  if  we  had  anything  from 
which  to  subtract. 

EXERCISES  IN  EQUATIONS. 

§  159.  The  learner  must  now  generalize  the  problems  in 
sections  3  and  4  of  Equations,  pages  34  and  39.  This  he 
can  easily  do,  if  he  takes  care  to  make  one  of  the  first  letters 
of  the  alphabet  stand  for  each  numeral  quantity;  say  a  for 
the  first  mentioned,  b  for  the  second,  &c.  When  the  same 
numeral  quantity  occurs  more  than  once  in  the  same  question, 
the  same  letter  must  stand  for  it  each  time. 


Questions.     How,  in  such  cases,  can  we  change  the  signs  1  How 

does  an  algebraical  sum  differ  from  an  arithmetical  sum?  How 

may  an  algebraical  sum  be  illustrated  in  common  language  1  What 
does  a  negative  quantity  standing  by  itself  denote  ? 


§161.]      MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.       117 


n. 

MULTIPLICATION    OF   ALGEBRAICAL 
QUANTITIES. 

§  160.  We  have  shown,  §39,  that  a  simple  literal  quan- 
tity may  be  multiplied  by  writing  the  multiplier  before  that 
quantity.  This  is  the  case  whether  the  multiplier  is  numeral 
or  literal.  Thus,  a  times  x,  is  written  ax;  bxc  =  bc.  In 
the  same  manner,  a  times  be  becomes  abc ;  and  f  times  abc 
becomes  fabc.  As  fxabc  is  the  same  as  abexf  we  see  that 
it  is  of  no  consequence  what  order  we  make  of  the  letters  in 
the  product,  abed  —  aedb  =  cadb,  &c.  But  it  is  generally 
more  convenient  to  follow  the  order  of  the  alphabet. 

Case  1. 

§  161.  Therefore,  to  multiply  one  simple  quantity  by  an- 
other, write  the  quantities  one  after  another,  without  any 
sign  between  them.  Thus,  abx  times  cfy  =  abxcfy;  5ax 
times  cdf—5acdfx.  But,  if  there  are  more  than  one  nume- 
ral co-efficient,  those  co-efficients  must  be  multiplied  as  in 
arithmetic,  and  placed  before  the  product  of  the  literal  quan- 
tities.    Thus,  da  x  2x  =  3.2ax  =  Qax.     2bc  X  5rs  =  1  Obcrs. 

EXAMPLES. 

1.  Multiply  a  by  b.  Ans.  ab. 

2.  Multiply  ab  by  c.  Ans.  abc, 

3.  Multiply  ab  by  cd.  Ans.  abed. 

4.  Multiply  2acx  by  by.  '  Ans.  2abcyx. 

5.  Multiply  3brs  by  2mnx.  Ans.  Qbmnrsx. 

Questions.  How  is  a  simple  literal  quantity  multiplied  1  In  what 
order  should  the  factors  be  written  1  What  is  the  operation  when 
there  are  numeral  co-efficients  1 


118  ALGEBRA.  [EQ.  SEC.  5. 

6.  Multiply  2adn  by  5cmx. 

7.  Multiply  Srsyop  by  4antx. 

8.  Multiply  2abcx  by  Sabrx. 

§  162-  By  the  foregoing  principle,  axa  =  aa.  Now,  as 
in  algebra  the  same  factor  is  often  found  two  or  more  times, 
Stifelius  (A.  D.  1554)  adopted  a  method  for  shortening  such 
expressions,  in  which  he  has  ever  since  been  followed.  The 
method  is  this :  when  the  same  letter  enters  as  a  factor  two 
or  more  times  into  any  quantity,  we  write  the  factor  but 
once,  and  put  at  the  right  of  it  and  a  little  raised,  a  figure 
denoting  how  many  times  it  has  been  multiplied.  Thus, 
aa  is  written  a8;  bbb  is  written  b3;  xxxx  is  written  x*; 
aabbbyyyy  is  written  a9  6s  y*. 

§163.  Mathematicians  are  accustomed  to  call  aa  or  a3, 
the  second  power  of  a,  or,  a-second  power;  a3,  is  called 
a-third  power,  &c. 

§  164,  The  figure  that  denotes  the  power  of  any  quantity 
is  called  the  exponent  or  index  of  that  quantity. 

§  165.  All  Quantities  are  said  to  have  an  exponent,  either 
expressed  or  understood.  Thus,  a  is  the  same  as  a1 ;  b=b1; 
&c.  The  written  exponent  affects  no  letter  except  the  one 
over  which  it  is  written ;  unless  it  is  denoted  by  a  vinculum. 

§  166.  Great  care  must  be  taken  by  the  pupil  not  to  con- 
found the  co-efficient  with  the  exponent;  as  their  effects  are 
entirely  different.  The  co-efficient  shows  addition,  the  ex 
ponent  denotes  multiplication.  For  example,  if  a  =  5,  then 
3a  =  5  +  5+5-  =  15;  but  a3  =  5x5x5 -  =  125. 

Questions.  About  what  time  did  Stifelius  write  ?  What  if  two  or 
more  factors  are  represented  by  the  same  letter?  How  are  such 
quantities  read  ?  What  effect  has  the  exponent  when  it  is  on  the 
right  of  several  letters'?  What  if  there  is  no  exponent  written  over  a 
letter?  Explain  the  difference  between  an  exponent  and  a  co-effi- 
cient, i 


$169.]    multiplication  of  algebraical  quantities.      119 

Case  2. 

§  167.  We  have  seen  that  a2  =  aa;  and  that  a3=aaa; 
now  aaxaaa  =  aaaaa.  So  we  see  that  a2xa3  =  a5.  Hence 
we  establish  the  rule  that  when  both  multiplier  and  multi- 
plicand are  denoted  by  the  same  letter,  their  product  is  found 
by  adding  their  exponents,     x2 x#4  =  xG,  y3 xy2  =  y5 ;  &c. 

§  168.  In  order  to  facilitate  the  practice  of  multiplication, 
it  is  best  to  observe  the  following  method :  First,  determine 
the  sign,  then  the  co-efficient,  then  the  letters  in  their  order, 
and  then  the  exponents. 

EXAMPLES. 

9.  Multiply  2amn  by  a.  Ans.  2a2mn. 

10.  Multiply  dabcx  by  4ax.  Ans.  12a2bcx». 

11.  Multiply  5bcmn  by  36c.  Ans.  \5b2c2mn. 

12.  Multiply  6a2xy  by  4ax2y.  Ans.  24a3x3y*. 

13.  Multiply  4a2bcx  by  a*b2c.  Ans.  4aeb3c2x. 

14.  Multiply  3a3m2  by  a*m3n. 

15.  Multiply  5a7 x6  by  4a3x4. 

16.  Multiply  2m5r2a2x  by  9a~x4rn*. 

17.  Multiply  Sb*c2n5  by  Sa3d2mn3. 

18.  Multiply  7a8y*x4  by  I2a5b2x. 

19.  Multiply  9m7x2a3  by  la2b3m. 

20.  Multiply  12«2m6na?5  by  18b3cnsy 

Case  3. 

§  169.  When  one  of  the  factors  is  a  compound  quantity, 
(§  65  and  67,)  we  multiply  each  term  by  the  other  factor,  and 

Questions.  How  is  multiplication  performed  when  there  are  ex- 
ponents? What  is  a  good  method  in  multiplication?  What  if  one 
of  the  factors  is  a  compound  quantity  % 


120  ALGEBRA.  [EQ.  SEC.  5. 

set  down  their  products,  each  with  their  proper  sign.    In  doing 
this,  we  generally  begin  at  the  left ;  thus, 

a+bc  ah — c. 

Multiplied  by  x  Multiplied  by  a 

ax-\-bcx  -  a~b — ac. 

21.  Multiply  ax-{-bx    by  xy.  Ans.  ax2y-\-bxsy. 

22.  Multiply  2a2bc —  rxy*  by  2ay.   Ans.  4a*bcy — 2arxy4. 

23.  Multiply  lay -fl     by  3r.  Ans.  21an/-f.3r. 

24.  Multiply  2xy  + ab+c*    by  2ax. 

25.  Multiply  5  —  7x+2a3b     by   4ac*y. 

26.  Multiply  3m3n  — 3rjo+2sa?3     by  6a*bd. 

27.  Multiply  2en  —  4an+ 5     by    12a9a?5. 

28.  Multiply  4y+y*    by  2rcy. 

29.  Multiply  3a  —  46+4     by    5y. 

30.  Multiply  6aa?8—  a3—  1    by   Sax9. 

31.  Multiply  4 -f  3a  —  a?9    by  ay. 

32.  Multiply  2a+569-f  3c  — 5e     by    3a8. 

33.  Multiply  769  — 4+a8#  — a:9    by    '4a»x. 

34.  Multiply  6x+7a—axy—2y     by    3bx*. 

Case  4. 
§  170.  As  (a+6)  x#  =  aa:+for;  we  know  that  x  x(a+b) 
also  =  ax+bx.     Whence  we  see  that  if  we  wish  to  multiply 
x  by  a+b ;  we  first  find  the  product  of  a  times  x,  and  then 
add  to  it  the  product  of  b  times  x.     And  generally,  when  the 
multiplier  and  the  multiplicand  is  each  composed  of  several 
terms,  the  product  is  made  up  of  the  sum  of  the  products  of 
the  multiplicand  by  each  term  of  the  multiplier. 
Thus,  (x+y)x{a+b) 
Cx+y     -J  Cx+y     ")        C  x+y 

=  <  a  y  and  <  b  V  or  <  a-f-o 

Lax+ayJ  Lbx+byj        Lax+ay+bx+by. 

Question.    How  is  multiplication  performed  when  there  are  seve- 
ral terms  in  the  multiplier? 


§  171.]    MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.  1<U 

EXAMPLES. 

35.  Multiply  2a+3b     by    4a+5b. 

2a +  36 
4a +  56 
8a2+\2ab 

+  10a6  +  15&3 
Uniting  terms,         8aa+  22a&  +  1565. 

36.  Multiply  a+6    by  a+6.  Ans.  a3+2a6+6» 

37.  Multiply  a +6  —  c     by    a2+6. 

Ans.  a3+a2b  —  a2c+ab+ba  —  be 

38.  Multiply  x2+2xy+y2    by   a?+i/. 

Ans.  x3+3x2y+3xy2+y* 

39.  Multiply  a— a?    by    2a+3x.      Ans.  2a2+ax  —  3x2 

40.  Multiply  a?  — 4     by    x+8.  Ans.  x2+4x  —  32 

41.  Multiply  a?  —  y     by    a?+y.  Ans.  x2  —  y2 

42 .  Multiply  4a2  —  3xy2+lax  —  ay     by    2a3 + ar2y. 

43.  Multiply  a2bc3  +  3a3c2  —  ob2c     by    ax2-\-yx2. 

44.  Multiply  m8a2+2rc2&3  — 1— m4n     by    2m862+4a8n*. 

45.  Multiply  8a2x3y*—2a4x3y2+a5y5     by    3a2m+2ny*. 

Case  5. 

§171.  As  (a  —  a?)  xb  =  ab  —  a$;  we  also  know  that  b 
x(a  —  x)=zab  —  xb.  Therefore,  to  find  a  —  x  times  b,  we 
first  find  the  product  of  a  times  b,  and  then  subtract  from  it  x 
times  b.  This  principle  is  very  plain  in  figures.  Thus,  sup- 
pose we  have  7  times  8,  and  wished  only  4  times  8 ;  if  we 
take  3  times  8  away  from  the  7  times  8,  we  should  obtain  4 
times  8 ;  as  56 — 24  =  32.  Whence  we  have  the  general  rule, 
if  there  is  a  negative  quantity  in  the  multiplier,  the  product 
of  that  quantity  and  the  multiplicand  must  be  subtracted 
from  the  product  of  the  positive  quantities  and  the  multi 
plicand.    H^"This  explains  the  rule  in  §  173. 

Question.     What  if  there  is  a  negative  term  in  the  multiplier? 
11 


122  ALGEBRA.  [EQ.  SEC.  5. 

Thus  (a  +  b)  x(c—d) 
a+b     "J        a+b 
minus  -Id  lor  c — d 


Lac 


id+bd)  ac+bc—ad—bd. 
Here  we  see  that  the  product  oft/ -into  a-\-b  is  subtracted 
from  the  product  of  c  into  a+b;  and  therefore  the  signs  of 
ad+bd  are  changed  to  — ad — bd. 

§  17.2-  By  examining  the  answer  of  this  last  example,  we 
shall  observe  a  principle  which  will  enable  us  to  be  more 
rapid  in  the  multiplying  operation.  It  is  this  :  when  we  mul- 
tiply a  +  term  by  a  +  term,  the  product  in  the  answer  is 
a  +  term  ;  and  when  we  multiply  a  +  term  by  a  —  term, 
the  product  in  the  final  answer  is  a  —  term.  It  will  be  well 
for  the  pupil  to  explain  this. 

By  understanding  this  principle,  we  are  able  to  set  the  final 
answer  down  at  first. 

EXAMPLES. 

46.  Multiply  a+b   by  a — 6. 
a+b 
a—b 
aa+ab 
—ab—ba 


a*         —  b*  Ans.  a*— 6s. 

47.  Multiply  x +Sz    by   3#9— 7xz. 

Ans.  3x*+l7x»z— 56xz\ 

48.  Multiply  x*+xy+y*    by   x—y,  Ans.  xs—y3. 

49.  Multiply  2x+3y  by  3x—4y.     Ans.  6x2+xy— 12y*. 

50.  Multiply  b3+ b*x+bx*+x*    by  b—x.     Ans.  b*—x*, 

51.  Multiply  a — b     by    c — d. 

a— b     ~\  fa—b      ~)        a—b 

c  I  minus  <  d  I  or  c — d 

ac — be  J  Lad — bd)      ac — be — ad+bd. 

Question.    How  is  the  subtraction  performed  1    By  what  principle 
«an  we  multiply  and  subtract  at  the  same  time  1 


§  173.]    MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.  123 

Here  we  see  that  in  subtracting  d  times  a — b,  we  change 
the  signs  of  ad — bd  to  — ad-j-bd. 

§173.  Whence  we  learn,  that  when  the  signs  are  alike, 
the  product  is  positive;  when  the  signs  are  unlike,  the  pro- 
duct is  negative  ;  or,  in  abbreviated  language, 
-f  multiplied  by  -f-,  produces  -f- 
4-  multiplied  by  — ,  produces  — 
—  multiplied  by  -f ,  produces  — 
• —  multiplied  by  — ,  produces  +• 
And  by  remembering  this  we  can  always  set  down  the 
final  answer  at  first. 

52.  Multiply  a—x  by  a— x. 

a  —x 
a  — x 
a2 — ax 
—  ax  -f-  a?8 


a3— 2ax+x*. 


53.  Multiply  2x— 3a    by   4x— 5a. 

Ans.  8x*— 22a#+15a3, 

54.  Multiply  2a  —  5y  by  a—2y.     Ans.  2a3—  9ay+l0y9. 

55.  Multiply  as-{-ac — c3    by   a — c.     Ans.  a3 — 2ac3-f  c8. 

56.  Multiply  a +b  —  d   by  a— b.  Ans.  a3— ad— b*+bd. 

57.  Multiply  Ax— 5a— 26     by  3#— 2a+56. 

Ans.  12x3— 23a#  -f  146#-f-  10a3— 21  ab  —  lOb*. 

58.  Multiply  x3 — y3 —  z3    by  x  —  y  —  z. 

59.  Multiply  6+ xy—  a3  —  my3     by   a3  —  3#3-fi/*. 

60.  Multiply  2a36  —  3ac3-f  463c3  —  1 

by  2a3c3-563c  —  8a3. 

61.  Multiply  3a3x3  —  4x5+263c3  —  abc—  1 

by  2#63+6c3#  —  6c—  1. 

62.  Multiply  Amx  —  nr  —  2 w2r+ 3 nr3— 1 

by  4ra3+3n3  — r3— 1 

Question.    What  is  the  general  rule  for  the  signs  ? 


124  ALGEBRA.  [EQ.  SEC.  5 

§  1 74.  The  rule  for  the  signs  which  has  been  established 
for  compound  quantities  is  also  extended  to  simple  quantities. 
It  may  seem  strange  to  the  young  student  to  hear  of  mul- 
tiplying -fa  by  — x,  or  —a  by  -}-#,  or  —a  by  — x.  But, 
as  it  was  intimated,  §157,  158,  a  negative  quantity  implies 
merely  a  contrary  value  from  a  positive  quantity  ;  and  there- 
fore, though  in  algebra,  the  operation  seems  to  be  upon  ima- 
ginary quantities,  yet  in  the  application  of  the  operation  to 
arithmetic  and  geometry,  the  mind  readily  understands  their 
signification.  The  extension  of  this  rule  to  simple  quantities, 
will  interpret  the  peculiar  results  to  which  algebraical  opera- 
tions sometimes  lead. 

§  175.  It  is  often  the  case  that  it  is  better  to  denote  the 
multiplication  of  compound  quantities,  than  to  perform  it,  on 
account  of  operations  that  follow.  Thus,  a — b  times  x-f  y, 
may  be  written  (a — b) .  (x+y).  No  general  rule  can  be  given 
to  determine  when  one  method  is  preferable  to  the  other. 
Experience  is  the  best  teacher  in  this  particular.  But  it  was 
thought  best  to  mention  it  in  this  place. 

When,  after  the  multiplication  has  been  denoted,  the  seve- 
ral terms  are  actually  multiplied,  the  expression  is  said  to  be 
expanded. 

§  176.  In  uniting  the  terms,  if  there  are  literal  co-efficients 
of  similar  quantities,  they  may  be  enclosed  in  a  parenthesis, 
as  was  shown  in  addition.     §  154  and  155. 

63.  Multiply  ma?3 — nx  —  r    by   nx  —  r. 

Ans.  mnx3 — (n9-f  mr)x2-\-r*. 

64.  Multiply  pxz+qx  —  y    by  mx  —  n. 

Ans.  mpx?+(mq —  np)x3 — (my-\-nq)x-\-ny. 

65.  Multiply  ax2  —  bx-\-c    by  x2  —  ca?+l. 

66.  Multiply^3 — rx-{-9    by   a?3  —  rx  —  q. 

Questions.  How  may  multiplication  be  denoted  %  In  what  cases 
may  literal  co-efficients  be  enclosed  in  a  parenthesis  ? 


§  179.]  GENERAL    PROPERTIES    OF    NUMBERS.  125 

in. 

GENERAL  PROPERTIES  OF  NUMBERS. 

§  177-  We  have  before  stated  that  algebraical  operations, 
(§  139,)  by  reason  of  the  quantities  themselves  being  retained 
in  their  original  value,  do  show  us,  in  their  results,  important 
general  principles.  We  will  here  make  a  few  multiplications 
of  some  quantities,  whose  results  show  us  some  remarkable 
general  properties  of  numbers.  These  properties  the  pupil 
should  remember,  as  they  are  of  frequent  use  in  the  subse- 
quent parts  of  this  study. 

§  178.  Suppose  we  have  two  numbers,  a  and  6,  of  which 
a  is  the  greatest.  Then  their  sum  =  a+6  ;  and  their  differ- 
ence m  a — 6.     Then  a  -f-  b 

a  — b 
a2  +  ab 
—  ab  —  b* 


a*  —  6s 

By  this  operation,  we  find  a  general  property  of  numbers 
which  it  would  be  difficult  to  find  by  any  arithmetical  opera- 
tion. It  is  that,  if  we  multiply  the  sum  of  two  numbers  by 
their  difference,  the  product  will  be  the  difference  of  the 
squares  of  those  numbers. 

§  179.  Again,  take  the  same  quantities,  and  multiply  their 
sum,  by  their  sum. 

a  +b 
a  +b 
a*  +  ab 
+  ab  -f  b* 


a*  -|-  2ab  -f-  b» 
By  this  operation  we  find  the  following  general  property 

Questions.    How  may  be  found  the  difference  of  the  squares  of  two 
quantities]     What  is  the  square  of  the  sum  of  two  numbers] 

11* 


126  ALGEBRA.  [EQ.  SEC.  5. 

The  square  of  the  sum  of  two  numbers  is  equal  to  the  square 
of  the  first  number,  plus  twice  the  product  of  the  two  num- 
bers, plus  the  square  of  the  last  number. 

§  180.  Again,  take  the  same  quantities,  and  multiply  their 
difference,  by  their  difference. 
a  —  b 
a  —  b 
a»  —  ab 
—  ab  +  b* 


a*  —  2ab  +  b» 
Therefore,  the  square  of  the  difference  of  two  numbers,  is 
equal  to  the  square  of  the  first  number,  minus  twice  the  pro- 
duct of  the  two  numbers,  plus  the  square  of  the  second. 

§  181.  The  only  difference  between  the  square  of  the  sum, 
and  the  square  of  the  difference,  is  in  the  second  term ;  being 
in  one,  positive,  and  in  the  other,  negative.  Let  us  subtract 
one  from  the  other. 

a*  +  2ab  +  b* 
a*  —  2ab  +  b*     ' 
4a~b 
The  actual  difference  between  the  square  of  the  sum,  and 
the  square  of  the  difference,  is  four  times  the  product  of  the 
two  numbers. 

§  182.  If  when  the  sum  of  two  quantities  has  been  raised 
to  the  second  power,  and  the  co-efficient  of  the  second  term 
has  been  rejected,  the  quantity  thus  obtained  be  multiplied  by 
the  difference  of  the  two  original  quantities;  the  result  will 
be  the  difference  of  the  third  powers  of  the  two  quantities. 
Also,  if  we  perform  the  same  operation  with  the  difference  of 

Questions.  What  is  the  square  of  the  difference  of  two  numbers? 
How  do  the  expressions  of  the  foregoing  squares  differ?  What  is 
their  actual  difference  in  quantities]  How  may  we  obtain  the  dif- 
ference of  the  third  powers  of  two  quantities?  How  may  we  obtain 
the  sum  of  the  third  powers  of  two  quantities  ? 


§  185.]    MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.  127 

the  two  quantities,  multiplying  by  their  sum,  we  shall  obtain 
the  sum  of  the  third  powers  of  the  two  quantities. 
Thus,  a  +  b  a  —  b 

a  +  b  a  —  b 

a*  +  2ab  -f  63  a*  —  2ab  +  b* 

Rejecting  the  co-efficients, 

a>  +  ab  +  b»  a*  —  ab  +  b* 

a  — b  a  +  b 


a3  +  a2b  +  ab»  a3  —  a2b  +  ab* 

—  a*b  —  ab»  —  b3  +  a2b  —  ab*  -f  b* 

a'  —  b3  a3  +  63 

FACTORS. 

§183.  In  all  simple  terms  consisting  of  more  than  one 
letter,  the  factors  are  evident  on  inspection.  The  same  is 
true  of  polynomials  that  have  a  simple  quantity  for  a  factor. 
Thus,  I0a'2b2x  +  \5a3bx  —  20a3b3y,  may  be  decomposed 
into  5a2b(2bx+3ax  —  4ab2y). 

§184.  In  order  to  ascertain  the  factors  of  any  quantity, 
first  see  what  is  the  greatest  quantity  that  will  divide  every 
term  of  the  given  quantity,  and  set  that  down  as  one  factor. 
Then  divide  the  given  quantity  by  the  factor  set  down,  and 
the  quotient  will  be  the  other  factor. 

§  185.  By  §178,  179,  a  +  b  is  a  factor  of  a»+2ab+b9; 
and  a  —  b  is  a  factor  of  a3  —  2ab+b*.  By  §177,  either 
a-J-6  or  a  —  b  is  a  factor  of  a8  —  63.  By  §182,  a+b  is  a 
factor  of  a3+£>3,  and  a  —  b  is  a  factor  of  a3 — b3. 

Any  quantity  is  a  factor  of  an  expression  made  up  of  the 
same  quantity  and  an  exponent. 

EXAMPLES. 

Find  the  factors  of  the  following. 

1.  4a3y— 4a*x+8a*xy—  12a7. 

2.  a3b2x*+3a*b*x3— a3a?a+4a7#3i/— 5a3x5y*. 

3.  2a3bx*—4ab4x  +  0a5b2x—2abx. 

4.  a9+2a6+63-f  a3— 63-fa3+63. 


128  ALGEBRA,  [EQ.  SEC.  5 

IV. 

DIVISION  OF  ALGEBRAICAL  QUANTITIES. 

§186.  Division  may  be  represented  by  the  sign  — ,  as 
a-i-by  is  read  a,  divided  by  b;  (a-f  b)  -f-  (c — d),  is  read 
a 4- 6,  divided  by  c — rf.  But  the  most  usual  way  to  denote 
division  (§  72)  is  to  write  the  divisor  underneath  the  dividend; 
.        a         a+b 

lhus-  v    z=* 

§187.  But  it  often  happens  (§76,  78)  that  the  fraction 
made  by  this  representation  is  an  improper  fraction;  and  one 
in  which  the  numerator  can  be  actually  divided  by  the  deno- 
minator. In  such  cases,  it  is  generally  best  to  perform  the 
division.     We  have  always  done  so  in  the  former  part  of  this 

treatise.     Thus,  4x  divided  by  2  equals  2x ;     — -  =  2x. 

o 

§  188.  Let  us  first  look  at  the  case  where  the  same  quan 
tity  is  in  both  the  dividend  and  the  divisor.  7a-i-7  =  a; 
126 -h  12  =  6.  In  the  same  manner,  ab-r-a  =  b;  dcs-d  =  c. 
This  may  easily  be  proved.  For,  ab  is  the  product  of  a  into 
b;  and  of  course  if  we  divide  by  what  was  the  multiplier,  we 
shall  obtain  the  old  multiplicand  again ;  as  may  be  seen  by 
trying  the  product  of  any  two  numbers. 

§  189.  Whence  we  derive  the  general  rule,  that  when  the 
divisor  is  found  as  a  factor  in  the  dividend,  the  division  is 
performed  by  erasing  that  factor  from  the  dividend, 
amn-r-a  =  mn,  because  amn  =  a  times  mn.  amn~m  =  an, 
because  amn  =  m  times  an.  amn-r-n  —  am,  because  amn 
is  n  times  am. 

Questions.  How  is  division  usually  represented  1  What  if  the 
same  quantity  is  in  the  dividend  and  in  the  divisor  1  Prove  it  by 
examples. 


§191.]  DIVISION  OF  ALGEBRAICAL  QUANTITIES.  129 

EXAMPLES. 

1.  Divide  8c   by  8.  Ans.  c. 

2.  Divide  be   by  b.  Ans.  c. 

3.  Divide  7ra   by  7.  Ans.  m. 

4.  Divide  am   by  a.  Ans.  m. 

5.  Divide  6rf   by  b. 

6.  Divide  7a6   by  7. 

7.  Divide  ca6   by  c. 

8.  Divide  6ad  by  b. 

9.  As  a&rf  is  the  same  product  as  the  last,  divide  that  by  b. 

Ans.  ad. 

10.  Divide  cde  by  c. 

11.  Divide  the  same  product  in  another  form,  thus,  dec 
by  d.  Ans.   ec, 

12.  Divide  «6c  by  c. 

13.  Divide  a&c   by  b. 

§  190.  As  a*  =  «a,  it  is  evident  that  if  we  divide  a2  by  a, 
the  quotient  is  a;  because  we  take  away  one  of  the  written 
a's.  So,  if  we  divide  a5  by  a, .the  quotient  is  a*;  because  as 
is  the  same  as  aaaaa,  which  -~  a,  gives  aaaa  or  a*.  So,  if 
we  divide  ah  by  a2,  the  quotient  is  aaa  or  a3.  And  if  we  divide 
a5  by  a4,  the  quotient  is  a;  because  aaaaa -h- aaaa  =  a. 
Hence  we  see,  that  when  there  are  exponents  in  either  the 
dividend  or  divisor,  the  division  is  performed  by  subtract- 
ing the  exponent  of  the  divisor  from  the  exponent  of  the 
dividend.     b5-.-b*  =  b3;    x7-r-x3  =  x*. 

§  19.1a  As  every  literal  quantity  is  understood  to  have  the 
number  1  for  its  co-efficient,  it  is  evident  that  if  we  divide  by 
1,  the  quotient  would  be  the  same  literal  quantity.  Thus, 
a-7-\  =  a.     And   again,   if  we   divide   a  literal   quantity  by 

Questions.  How  is  division  performed  when  there  are  exponents? 
Why  ]  How  when  the  divisor  is  the  only  quantity  in  the  dividend? 
Why? 

I 


130  ALGEBRA.  [EQ.  SEC.  5. 

itself,  the  quotient  will  be  1.     Thus,  la—a  =  1 :  the  divisor 
being  a  factor  in  the  dividend. 

§  192.  It  sometimes  happens  that  the  co-efficient  contains 
the  divisor  as  a  factor.  Thus,  8a  is  the  same  as  2  times  4a, 
or  4  times  2a ;  and  therefore  can  be  divided  by  2  or  by  4. 
In  the  same  manner  Sab  may  be  divided  by  2a,  or  4a,  or  26, 
or  4b  ;  because  it  is  2a  times  4b,  or  4a  times  2b. 

We  have  only  to  remember  to  take  those  factors  out  of  the 
dividend  which  are  equal  to  the  divisor.  And  in  general, 
when  there  are  co-efficients  in  both  the  divisor  and  the  divi- 
dend, divide  the  co-efficient  of  the  dividend  by  the  co-efficient 
of  the  divisor;  and  then  proceed  with  the  literal  quantities  as 
before  directed.     I0abc-r-5b  =  2ac;    I2a3xy-r-3a2y  —  4ax. 

EXAMPLES. 

14.  Divide  a3  by  a9.  Ans.  a. 

15.  Divide  x6   by  x*.  Ans.  x*. 
10.  Divide  a3b*y   by  a*b.  Ans.  ab8y. 

17.  Divide  d3c*x7   by  rfc3x*.  Ans.  d2czx*. 

18.  Divide  a*m*x  by  a*m*. 

19.  Divide   a*xay7   by  ax4y*. 

20.  Divide  d6y7   by  dy. 

21.  Divide  p3r7sat  by  r5st. 

22.  Divide  ab3cAd*  by  ab»cd5. 

23.  Divide  ax7ya   by  ax?ys. 

24.  Divide  cbr7sHx*y7  by  c*r5ty*. 

25.  Divide  6a*bc7   b^  26c5.  Ans.  3a9c". 

26.  Divide  \2ax*y*  by  4a?/9.  Ans.  3x2ya. 

27.  Divide  2\bc3xyR   by  3ct/3. 

28.  Divide  42c7rfV}   by  6c2d5x. 


Questions.   How  is  division  performed  when  there  are  co-efficients  ? 
Explain. 


§   195.]  DIVISION  OF  ALGEBRAICAL  QUANTITIES.  131 

29.  Divide  36pr*st*   by  4rsf. 

30.  Divide  54m5n2x32/   by  9m3ny. 

31.  Divide   66a*c8aV   by  3a*csa?. 

32.  Divide  48c*r7x3y5  by  8cr5xy*. 

33.  Divide  72aV?»4  by  18a2rm3. 

Case  2. 

§193.  We  have  shown,  §169,  that  (a+b)xc  =  ac+bc. 
Of  course  (ac  +  bc)-±-c  =  a-\-b;  where  we  see  that  ivhen  we 
divide  a  compound  quantity  by  a  simple  quantity,  we  divide 
each  of  the  terms  by  that  quantity. 

§194.  We  must  also  recollect  that,  as  -f  multiplied  by 
-f ,  makes  +  in  the  product,  so  -f  in  the  product  divided  by 
-f »  must  make  -f  in  the  quotient.  And  that  as  -f-  multiplied 
by  —  makes  — ,  so  —  in  the  product  divided  by  — ,  will 
bring  back  the  -f  in  the  quotient.  So  that  when  the  signs 
are  alike  in  the  dividend  and  divisor,  the  sign  in  the  quo- 
tient is  +.  Thus,  axb  =  ab;  both  of  which  are  +•  Also 
— «X — b  =  ab;  and  — ab-. b=+a. 

§195.  Again,  as  —  multiplied  by  -f  makes  — ,  so  in 
the  product,  —  divided  by  -f-»  brings  back  —  in  the  quotient. 
Also,  —  multiplied  by  —  makes  -J-  ;  and  of  course,  +  in  the 
product  divided  by  — ,  brings  —  in  the  quotient.  That  is, 
when  the  signs  In  the  divisor  and  dividend  are  unlike,  the 
zign  in  the  qudient  is  — . 

examples. 

34.  Divide  2ad+8a»c   by  2a.  Ans.  d+4ac. 

35.  Divide  8tf*3m2—  I2d5m?  by  4dm9.  Ans.  2d»—3d*m 

36.  Divide  ixy  +  Gx*  by  2a?.  Ans.  2y-\-3x. 

37.  Divide  abc — acd  by  ac.  Ans.  b — d. 

38.  Divide  I2ax—8ab   by  —4a.  Ans.  —  3x+2b. 


Questions.     How  is  a  compound  quantity  divided  ?   What  are  the 
rules  for  the  signs  ?     For  what  reason  ? 


132  ALGEBRA.  [EQ.  SEC.  5. 

39.  Divide  \0xz  +  \5xy  by  5a?.     , 

40.  Divide    I5ax— 27a?   by  3a:. 

41.  Divide  18a-2— 9a?  by  9a?. 

42.  Divide  abc—bcd—bcx  by  —be. 

43.  Divide  3a?+6a?--f3aa?—  15a?   by  3a;. 

44.  Divide  3abc  +  \2abx— 9a»b   by  Sab. 

45.  Divide  40a362-f60aa&8— 17 ob   by  ab. 

46.  Divide   \5a-bc  —  l0acx*  +  5ad*c   by  —  Sac. 

47.  Divide  20ax+\5ax*  +  \0ax—  5a   by  5a. 

§  196.  It  is  evident  that  we  may  divide  by  either  factor 
Thus,  ax+bx  may  be  divided  by  x,  and  the  quotient  will  be 
a  +  b;  or  it  may  be  divided  by  a  +  b,  and  the  quotient  will 
be  x.  This  may  appear  singular  to  the  young  pupil ;  but  he 
is  to  recollect  that  division  is  merely  separating  the  dividend 
into  factors,  being  careful  to  make  one  of  them  of  a  given 
magnitude ;  that  is,  to  make  it  the  same  as  the  given  divisor. 
For  illustration,  ax+bx  means  that  x  is  taken  a  times,  and 
also  b  times.  Therefore  it  is  taken  (a  +  b)  times;  and  in  the 
whole  quantity,  a  +  b  is  the  co-efficient  of  a?,  so  that  (ax+bx) 
=  (a  +  b)x. 

§197.  Now  we  know  that  x  times  a+b  =  ax  +  bx;  and 
also  that  a+b  times  x  =  ax+bx.  Whence,  the  product 
(ax  +  bx)-i-(d  +  b)z=x.  Therefore  we  conclude  that  if  the 
divisor  contains  just  as  many  terms  as  the  dividend,  with 
corresponding  signs  ;  and  the  first  term  of  it  is  a  factor  in 
the  first  term  of  the  dividend,  the  second  term  of  it  in  the 
second  >>f  the  dividend,  a.i<!  so  on  through  <ach  of  them  re- 
spectively; and  the  remaining  factor  in  every  term  of  the 
dividend being  the  same  ;  thai  remaining  factor  only  shall  be 
the  quotient. 


Questions.  What  is  the  ruin  for  dividing  when  the  number  of 
terms  in  the  dividend  and  divisor  is  the  samel  Explain  and  illus- 
trate? 


§   199.]  DIVISION   OF  ALGEBRAICAL  QUANTITIES.  133 

EXAMPLES. 

48.  Divide  ax+bx+cx   by  a+b-\-c.  Ans.  x. 

i9.  Divide  bac+bc2x — bx3  by  ac+c2x — x3.         Ans.  b. 

50.  Divide  c2ax — 2abx — 3xy-\-x   by  ac2 — 2ab — 3y-\-l. 

51.  Divide  c^x—abd'  +  ^x3—,!2  by  ex— ab  +  x3—  1. 

52.  Divide  a2y — bcy-\-xy   by  a2 — bc-\-x. 

53.  Divide  6ah?n — \iabm — 3cdm   by  Qah — \4ab — 3cd. 

§  198.  If  the  letters  of  the  divisor  are  not  found  in  the 
dividend,  the  division  is  expressed,  as  we  have  before  shown, 
§  186,  by  writing  the  divisor  underneath  the  dividend,  in  the 
form  of  a  vulgar  fraction. 

EXAMPLES. 

4t/  -4-  73? 

54.  Divide  4t/-f7a?   by  a— b.  Ans.  -^—  ,— . 

55.  Divide  3a  +  2b2— c   by  a+c. 

56.  Divide  a3—x2b  +  c5  by  a3— b*. 

57.  Divide  3a2c+263+c   by  2c. 

§199.  When  the  dividend  is  a  compound  quantity,  the 
divisor  may  be  placed  underneath  the  whole  dividend  if  we 
choose.  It  may  also  be  placed  under  each  term  of  the  di- 
vidend, which  is  the  same  as  dividing  each  term,  according  to 
§  193.     By  this  method,  the  answer  of  the  last  sum  would  be 

3ac9      263        c  ,     -  n      .      ,     ,     ,  .       ,     , 

f- r-  — .     Answer  the  following  by  both  methods. 

2c         2c       Zc  , 

58.  Divide  3ab+b  +  2ab   by  a.  Ans. 1 1 . 

'  a     a        a 

59.  Divide  6a  +  ab— 3b   by  2b. 

60.  Divide  2x+2y-\-3ax—2a2y   by  3ay. 

61.  Divide  ax2—bx—a2b2-\-ab   by  2b. 

Questions.  What  is  the  operation  when  the  same  quantities  are 
not  in  both  the  dividend  and  divisor?  Must  the  divisor  always  be 
put  under  the  whole  dividend  1 

12 


134  ALGEBRA.  £eQ,.  SEC.  5. 

§  200.  When  we  divide  each  term  separately,  we  may  use 
both  methods  of  division ;  that  is,  we  may  actually  divide 
such  terms  as  we  can  by  §  189 ;  and  merely  express  the  di- 
vision in  such  terms  as  cannot  be  divided. 


EXAMPLES. 

ttX 

62.  Divide  cd — ax-\-ac+bc   by  c.     Ans.  d \-a+b 

63.  Divide  ax+bx— 2ab-\-2x  by  x.   Ans.a+6 f-2. 

x 

64.  Divide  2am— 3a*b+b*m— 3a*m  by  —a. 

bnm 

Ans.  — 2m-\-3ab ; \-3am. 

a 

65.  Divide  2b9— a*b+3b*c+a*b3— ac  by  b*. 

66.  Divide  ay*—by*+4a3b»—&a*by*+aby  by  ay. 

67.  Divide  abx*+a?by—3ab2x+ax*—la2y   by  ab. 

68.  Divide  2abm+6a*b  +  5b*m— 4a8m  by  ab. 

69.  Divide  by — a3b9y-\-ay — aby+bsy  by  — by.     • 

70.  Divide  ax — bx*-\-xy*+by — ay  by  — y. 


EXERCISES  IN  EQUATIONS. 

§  201.  The  learner  must  now  generalize  the  problems  in 
section  5  of  Equations,  page  44.  And  with  this  section,  we 
will  make  our  calculations  more  purely  algebraical  than  in  the 
preceding  sections ;  as  we  shall  take  care  to  use  no  numeral 
quantities  at  all  in  stating  the  questions. 

§202.  It  is  customary  to  represent  those  numbers  which 
stand  for  times,  by  the  letters  m,  n,  p,  a,  Sic. 

1.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money  in 
trade.  A  gains  c  ($126,)  and  B  loses  d  ($87 ;)  and  now  A's 
money  is  m  (two)  times  as  much  as  B's.  What  did  each  lay 
out  ?     See  page  44. 

Question.     When  may  we  use  both  methods  of  dividingr? 


§  202.]  DIVISION  OF  ALGEBRAICAL  QUANTITIES.  135 

Let     x  =  what  each  lay  out. 

x-\-c—  A's  sum  now. 

x — rf=  B's  sum  now. 

mx — md  =  m  times  B's. 

Forming  the  equation,  mx-—md  —  x+c        f 

Trans  posing,  mx — x  =  c  -f  md 

_...,.      ,  ,'•■--  c+md 

Dividing  by  m—  1,  x  = -. 

to    f  m—l 

Substituting  numbers   £  c  +  md      126  +  2x87 

r         1  f  7"  ==s 7^. 1 "  ^=  "v)U. 

for  letters,  )  m — 1  2 — 1 

In  sums  of  this  kind,  the  only  difficulty  is  to  determine  what 
quantity  we  must  divide  by  in  the  last  step,  to  leave  x  alone. 
But  this  difficulty  is  easily  overcome,  by  dividing  mentally 
the  left  hand  member  by  a?,  and  observing  the  quotient.  Of 
course,  dividing  the  same  member  by  that  quotient  will  pro- 
duce x;  which  is  our  only  object. 

2.  The  2d  sum  on  page  45,  is  performed  as  follows : 

x  =  the  wife's  age. 
mx  =  the  man's  age. 
x-\-a=  wife's  after  a  years. 
m#-fa=  man's  after  a  years. 
nx+na  =  n  times  the  wife's  age. 
Forming  the  equation,  mx+a  —  nx-\-na 

Transposing  terms,  mx — nx  =  na — a 

Dividing  by  m — n,  x  = . 

Substituting  numbers,  x  =  15,  the  age  of  his  wife. 

All  the  other  questions  in  section  5th  are  to  be  performed 
in  the  foregoing  manner. 


136  ALGEBRA.  [EQ.  SEC.  5. 


DIVISION  BY  COMPOUND  DIVISORS. 

§203.  By  examining  any  sum  in  multiplication,  it  will  be 
seen  that  any  one  term  in  the  product  contains  both  one  term 
of  the  multiplicand,  and  one  term  of  the  multiplier.  See  each 
term  in  the  answer  of  the  following  example. 

x+y 

a+b 


ax+ay+bx+by. 

Therefore,  if  we  find  any  term  of  the  product  that  contains 
the  first  term  of  the  multiplier,  we  know  that  the  remaining 
part  of  it  is  a  term  in  the  multiplicand.  And,  as  either  term  of 
the  multiplicand  may  be  put  as  the  first,  (for  y-\-x  is  the  same 
as  x+y),  we  may  suppose  that  the  term  which  we  have  found 
by  this  process  is  the  first.  Thus,  in  the  answer  above,  ax 
and  ay  contain  a  of  the  multiplier ;  and  therefore  either  x  or 
y,  whichever  we  choose,  is  the  first  term  of  the  multiplicand. 

Now,  by  change  of  names,  ax  +  ay+bx+by  may  be  the 
dividend,  and  a -\-b  the  divisor,  for  the  purpose  of  finding  the 
quotient.  We  see  that  a,  the  first  term  of  the  divisor,  is  found 
in  ax  and  in  ay ;  and  we  may  therefore  conclude  that  the  first 
terra  in  the  quotient  is  either  x  or  y,  which  are  the  remaining 
parts  of  those  terms.  Likewise,  {a -\-b)  x  x  or  y,  will  be  a 
partial  dividend. 

Ex.  1.  Suppose  it  were  required  to  divide  ac+bc+ad+bd 
by  a+b.  We  see  that  the  first  term  in  the  divisor  is  contained 
in  the  first  term  of  the  dividend ;  and  therefore  judge  that  the 
other  factor,  which  is  c,  must  be  the  first  term  in  the  quotient; 
and  that  c  times  a+b,  which  is  ac  +  bc,  is  one  of  the  partial 

Questions.  What  may  be  said  of  each  term  in  a  product?  What 
then  do  we  know  of  any  term  in  the  product  that  contains  the  first 
term  of  the  multiplier?  What  is  the  change  of  names  for  division 
;nstead  of  multiplication  ?  Explain  the  first  example.  To  what  is 
:t  similar  1 


§  205.]  DIVISION    BY    COMPOUND    DIVISORS.  137 

dividends.  Subtracting  ac-\-bc  from  the  whole  dividend,  and 
there  remains  ad-\-bd  still  to  be  divided  by  a-{-b.  The  work 
is  as  follows  : 

a-\-b)  ac-\-bc+ad-\-bd  (c+d 
-  ac+bc 

ad+bd 
ad+bd 

In  the  last  partial  dividend,  we  see  by  the  first  term  that  the 
divisor  is  contained  in  it  d  times. 

§  204.  This  is  the  same  method  that  is  followed  in  arith- 
metic ;  where,  in  order  to  find  a  quotient  figure,  we  see  how 
many  times  the  first  figure  of  the  divisor  is  contained  in  the  first 
figure  of  the  partial  dividend ;  and  supposing  the  number  to  be 
the  true  quotient,  we  multiply  the  divisor  by  it.  But  there  is 
this  difference  between  algebraic  and  arithmetical  division, 
that  in  algebra  the  quotient  is  always  the  true  one ;  and  also 
where  the  divisor  is  not  found  as  a  factor  of  one  of  the  terms 
of  the  dividend,  the  operation  of  division  is  impossible. 

Ex.  2.  Divide  5a8#-f  x3-\- a3+5ax*  by  a+x. 

§  205-  From  the  last  example,  it  seems  that  it  is  best  to 
put  first  in  the  dividend  those  terms  that  contain  the  letter 
which  is  in  the  first  term  of  the  divisor.  It  has  also  been 
found  easier  in  practice  to  arrange  the  powers  of  this  letter 
both  in  the  divisor  and  dividend,  so  that  the  highest  should 
stand  first,  the  next  highest  next,  and  so  on.  We  will  adopt 
this  plan  in  the  present  example. 

a+x)  a8-|-5a8a?-f5aa?a+ic3  (a*+4ax+x* 
In  subtracting,  a3-\-a*x 

change  the  signs         4a3.r-f-5aa,3-fa?8 

mentally,      and        4a»x+4ax*  ^j*  We  find  the 

unite      at      the  ax2+x*         terms    0f    qUOtient 

same  time.  ax*+x*         by  §  2Q4      ^^ 

The  first  term  of  the  divisor  is  contained  in  a3,  a8  times ; 

Questions.  To  what  is  this  method  similar  1  What  arrangement 
of  terms  is  recommended  ?     Explain  the  second  example. 

12* 


138  ALGEBRA.  [EQ.  SEC.  5. 

wherefore  we  determine  that  aa  is  one  term  in  the  quotient. 
The  product  of  that  term  and  the  divisor  is  a?  +  a2x,  which 
we  subtract  from  the  dividend,  and  there  is  left  4a2x-\-5ax9 
-f-cc3.  The  first  term  of  the  divisor  is  contained  in  the  first 
term  of  this  partial  dividend  4ax  time's ;  wherefore  we  deter- 
mine, that  4ax  is  one  term  of  the  quotient. 

We  will  incorporate  the  foregoing  remarks  into  the  follow- 
ing general  rule. 

RULE    FOR   DIVIDING    WITH    COMPOUND    DIVISORS. 

§  206.  Arrange  the  terms  of  both  dividend  and  divisor 
according  to  the  powers  of  the  same  letter,  beginning  with 
the  highest.  Divide  the  first  term  of  the  dividend  by  the 
first  term  of  the  divisor,  and  place  the  result  with  its  pro- 
per  sign  for  the  first  term  in  the  quotient. 

§  207.  Multiply  the  whole  divisor  by  this  term  of  the 
quotient,  setting  the  product  under  the  dividend;  subtract 
this  product  from  the  dividend,  and  use  the  remainder  as 
the  next  partial  dividend. 

§  208.  Divide  the  first  term  of  the  partial  dividend  by 
the  first  term  in  the  divisor,  and  place  the  result  for  the 
next  term  in  the  quotient.  MuViply  the  divisor  by  the  term 
last  written  in  the  quotient,  subtract  the  product  as  before, 
and  proceed  in  this  manner  as  long  as  the  first  term  of  the 
dividend  can  be  divided  by  the  first  term  in  the  divisor. 

§209.  If  there  is  a  remainder  in  which  there  is  no  term 
that  contains  the  first  term  of  the  divisor,  that  remainder 
must  be  made  the  numerator  of  a  fraction  whose  denomina- 
tor is  the  divisor;  and  the  fraction  must  be  put  at  the  end 
)f  the  quotient  by  addition. 

Questions.  What  is  the  first  step  in  the  rule  ?  What  is  done  with 
the  first  term  of  the  quotient?  How  do  we  proceed  with  a  partial 
dividend]     What  is  done  with  the  final  remainder? 


§210.]  DIVISION    BY    COMPOUND    DIVISORS.  139 

EXAMPLES. 

3.  Divide  a*+3a2x+3ax2+x3   by  a -fa?. 


a+x)  a3+3a9a?-f-3aa?2+a?3  (a3-f2aa?+a?3 
cf+a2x 

2a'x+3ax2+x3 
2a2x+2axa 


ax*+ x3 
ax2+xz 


The  first  term 
of  the  divisor  is 
contained  in  the 
first  term  of  the 
dividend  a2  times. 
The  whole  divisor 
multiplied  by  a3,  equals  a?-{-a2x;  which  subtracted  from  the 
dividend  leaves  2aix-\-Sax2-\-x3.  The  first  term  of  the  divi- 
sor is  contained  in  the  first  term  of  the  remainder,  2ax  times. 
The  whole  divisor  multiplied  by  2ax,  equals,  &c. 

§  210.  It  may  be  well  to  mention  here,  that  if  the  signs 
of  the  divisor  are  changed,  we  shall  obtain  the  same  quantities 
for  the  quotient  as  before,  with  the  exception  that  their  signs 
will  be  changed  also.  The  pupil  may  divide  this  last  sum 
by  — a — x ;  and  he  will  obtain  — a* — 2ax — a?8. 

4.  Divide  a?3— 6a?— 9   by  a?— 3.  For  subtracting  see  Ex.  2. 
x—3)  #3_6a?— 9  (a?3-f  3a?-f  3 
a?3— 3a?s 


+  3a?3—  6a?— 9 
3a?3— 9a? 

3a?— 9 
3x— 9 


5.  Divide  cr'-f  4aa?9—  3a3a?— 2a?3   by  aa— 2aa?-f  a?9. 

fl?j?2— a?3 
a»_ 2aa?+a?3)  a3— 3a2x+4ax2— 2a?3  (a—x+- 


a>-2a2x+ax2  a3-2aa?+a?9 

,      —  aaa?-f3a2?8— 2a?3 
— a3a?-r-2ga-a— a?3 
ax2 — 3? 

6.  Divide  a3— 3a2y+ 3ay*— y3  by  a— y. 

Ans.  a3— 2ay+y* 

7.  Divide  63— 10&3+33&— 36  by  6— 4.    Ans.  &*_6&-f9. 

8.  Divide  6a*— 96  by  3a— 6.       Ans.  2a3+4a8-f  8a+16. 


140  ALGEBRA.  [EQ.  SEC.  5. 

9.  Divide  ar3— 3a?86  +  3xb*—bs  by  x—b. 

Ans.  a?2 — 2xb-\-b2. 

10.  Divide  x* — y*  by  x—y.        Ans.  x^+xfy+xif-j-y*, 

11.  Divide  a* — 2a?*  by  a -fa?.  ^ 

Ans.  a^a^+a*2-*3-  — . 
a-fa? 

2#3 

12.  Divide  a3— 2?  by  a-fa?.       Ans.  a2 — aa?-f  a?2 


a+a? 

13.  Divide  x*+y*  by  a?-fy.  2w* 

Ans.  or1— tfy+xy1— ^+_iL-. 

14.  Divide  21/3— 19y2+26y— 16   by  y— 8. 

Ans.  2y2—3y+2. 

15.  Divide  48a?3— 76aa?8— 64a8a?-f  105a3  by  2a?— 3a. 

Ans.  24a?8— 2aa?— 35aa. 

16.  Divide  b*—3y*  by  b—y.  „  4 

Ans.  b'+b^+by'+tf-.^JL-. 

17.  Divide  2a4—  13a36+31a96«— 38a63-f  246* 

by  2afl— 3ab+4b*.  Ans.  a3— 5a&+6&8. 

18.  Divide  a?8+/>a?+ q  by  a?+ a.  a— pa-f-a8 

Ans.  a?-f»— a-f- — *■ . 

a?-f-a 

19.  Divide  6a?* -f 9a?9— 20a?  by  3a?8— 3ar.  _ 

Ans.  2ar8+2a?-f  5- 


3a?9— 3a? 

20.  Divide  9a?8— 46a?5  -f  95a?9 -f  150a?  by  a?8— 4a?— 5. 

Ans.  9a?*— 10a?8 + 5a?9— 30a?. 

3  77         43  33 

21.  Divide  -Xs— 4a?*+—  a?8 -a?8—  —a?-f27 

4  8  4  4 

1  3  1 

by  -a?9— a?-f  3.  Ans.  -a?8— 5a?9+-a?-f  9. 

2  Z  4 

§211.  It  sometimes  happens,  that  in  the  dividend  there 
are  several  terms  which  have  the  same  exponen.  ^f  that  letter 
by  which  the  quantity  has  been  arranged.  In  such  cases, 
they  should  be  put  directly  under  one  another ;  and  when  they 
become  the  first  of  a  partial  dividend,  there  should  be  a  quo- 
tient for  each  term  before  the  subtraction  is  performed;  as 
follows : 


§212.]  DIVISION    BY    COMPOUND    DIVISORS.  141 

22.    Divide    1  Oa3  +  1 1  a*b  -f-  3ab2  —  1 5a  c  —  5b2c  —  1 9a6c 
-j-l5bc*   by  5a2 -f  3a&  —  56c. 

*  2  ,  o  *    ^  O0a3+lla26-f  3a&2— 5&3c-fl5&c270      ,     0 
5aW_56c{        I15a2c_19a,c  }2«+6-3c 

10a3+  6a  6—  \0abc 

+  5a26  +  3a63— 563c  + 1 56c3 


f      -f   5a*&  +   'dab2— bb»c+lbbc2l 
1      —  15a3c—  9a6c  * 


Product  by  +6  =  5a26  -f  3a62— 563c 

Prod,  by—  3c  =      — 15a2c—  9a6c  +156c8 


0 


23.  Divide  a*+4a*b+8b*  by  a+2&.  \eb*+24b* 

Ans.  a3— 2a26+4a&-|-4aZ>2— 8&2— 8&3-' 


a +2b 

24.  Divide  64a3  +  64a&-fl6&8—  9a*3— 48a*-64 

by  8a  +  4&-f  3rf+8.  Ans.  8a  +  46— 3d—  8. 

25.  Divide  18a2  +  33a6-f  42ac— 1 2ae?— 30&3  +  1246c  +  86a* 
—  16c3— 32cd  by  6a+15&— 2  c— 4a*.  Ans.  3a-26+8c. 

26.  Divide  1  by  1—  x. 

l—x)  1       (\+x+x»+x*+x*,  &c. 

1—  X 
X 

x—x* 


a?3— x8 


x* 

aax9 


27.  Divide  a3  by  l—x8.       Ans.  a3-f-a3#3-f  a3a?6+ 


1— Xs 


§212.  In  the  last  two  examples,  the  division  may  be 
carried  on  forever,  like  a  decimal  fraction.  Any  algebraical 
fraction  also  may  be  expanded  by  actually  performing-  the 
division  in  this  manner.  But  as  in  this  example,  so  in  many 
others,  a  few  leadin/  terms  of  the  quotient  will  be  sufficient 
to  indicate  the  rest,  without  continuing  the  operation. 


142  ALGEBRA.  [EQ.  SEC.  5 

28.  Expand  the  fraction . 

r  a+x 

„       XX2      X3      X*     e 

«+*)*      (1- -+-_-+-,  to. 

a+x 
— x 

— a: IC7*  For  — --r-a  = 

a  la 


x%  x*       __x» 

a  a    '         a* 

x9  x*  xs       _      x* 

a  a*  ""as"^a  =  "a"9 


It  is  obvious  that  if  the  division  is  continued,  the  remaining 
terms  will  be  alternately  —  and  -f-  ;  and  that  they  will  in- 
crease one  power  in  every  successive  term. 

§  213.  Such  operations  may  be  carried  on  infinitely,  so  as 
to  bring  the  result  nearer  and  nearer  to  the  true  quotient. 
And  on  this  account  mathematicians  have  called  such  ex- 
pressions Infinite  series.     It  must  be  understood,  however, 

that  it  is  not  the  quantity  which  is  infinite,  but  the  number 

2 
of  terms.     Thus,  .66666666G66,  &c,  =  -. 

3 

x     x%     x3    x*     x5    x9    0  a 

1 1 1 — ,  &c.,  = . 

a^a*     a3^a*     a^  a6         '      a+x 

29.  Reduce to  an  infinite  series. 

\+a 


\+a 

30.  Reduce  , to  an  infinite  series. 

I— a 


Ans.  1— a  +  a2— a3+a4— a5+a6,  &c 

an  infinite  series. 
Ans.   l+2a  +  2a*+2as  +  2a*  +  2a5,  &c. 


Questions.  What  may  be  done  with  an  algebraical  fraction?  Is 
it  always  necessary  to  continue  the  operation?  How  far  may  the 
operation  be  continued  ?  What  name  has  been  given  to  the  answer? 
Is  the  quantity  infinite  ? 


§214."]  FRACTIONS.  143 

V. 

FRACTIONS. 
REDUCTION  OF  FRACTIONS  TO  LOWER  TERMS. 

§214.  We  showed  in  §88,  that  a  fraction  may  be  re- 
duced to  lower  terms,  without  any  alteration  in  its  value,  by 
simply  dividing  both  terms  by  a  number  that  will  divide  each 
without  a  remainder.  Fractions  that  are  expressed  by  literal 
quantities  may  frequently  be  reduced  in  the  same  manner. 

Thus,  in  the  fraction  — ^,  both  the  ierms  may  be  divided  by 

a;  and  the  fraction  will  then  become  -^-. 

o 

EXAMPLES. 

4abcx 


Reduce  to  the  lowest  terms  the  fraction 


Qadcy 

_    ,  C4abcx     „  2bx 

Ans.  Both  terms  <- — ; — i-2ac  =  —1-. 

C  Qadcy  3 ay 

■■    _   -         a2m3v        .     .  .        my 

2.  Reduce  —r — £-  to  its  lowest  terms.  Ans.  — £-. 

tfom^x  abx 

n    n   ,  56^4v8        .     ,  8v3 

3.  Reduce  — =-r~*  t0  lts  lowest  terms.  Ans. £_. 

— lxby*  x4, 

.    _    ,         —  4x5y*z       .     .  4x3z 

4.  Reduce  — -~ —  to  its  lowest  terms.         Ans. . 

5x  y*  by2 

m    ~    ,         —I2x4yz       .     .  3a? 

5.  Reduce  — —  to  its  lowest  terms.  Ans.  — . 

—4x*yz*  z% 

10°*  These  examples  will  remind  the  pupil,  that,  (because 
fractions   are   merely  expressions  of  division,)  when   each 


Questions.  How  may  fractions  be  reduced .  to  lower  terms  ? 
Where  must  a  sign  be  put  to  affect  the  whole  fraction  1  What  sign 
in  e  tch  term  will  make  plus  for  the  whole  1  What  sign  in  each  term 
will  make  miniu  for  the  whole  ?    §  195. 


i44  ALGEBRA.  [EQ.  SEC.  6 

term  has  its  sign,  then  the  whole  fraction  will  have  a  sign 
according  to  §  194  and  195. 

6.  Reduce     •     • —  to  its  lowest  terms. 

I4ab*x* 

9x3y5 

7.  Reduce  — — —  to  its  lowest  terms. 

45ay*  , 

\5(t3r9s 

8.  Reduce  —  to  its  lowest  terms. 

27amst 

§  215.  When  we  divide  a  compound  quantity  by  a  simple 
quantity,  we  divide  each  term,  §193.  Hence,  in  reducing 
fractions  to  lower  terms,  we  must  find  for  the  divisor  a  quan- 
tity that  is  a  factor  in  every  term,  both  of  the  numerator  and 
denominator. 

EXAMPLES. 

«    t>  j        a*x+ay3  ,  ax-\-y* 

9.  Reduce  --*-   to  its  lowest  terms.        Ans.  V  -  • 

a*b  a»b 

m    t>   i        a*x»+ax9—3a*x       .     . 

10.  Reduce       r— -  -    -     to  its  lowest  terms. 

ax*— 6ax'+9a8x  *   o 

ax2  +  x  —  3a8 
Ans.  — - — -- — — . 
x3— 6x-f  9a 

11.  Reduce  — ^ — - — — - — 2  to  its  lowest  terms. 

8axJ  —  4a*x 

.«    n  j        3a3b*x  —  9b  x  -6axa       .     , 

12.  Reduce  — - — — — — z —  to  its  lowest  terms. 

I2ax+6abx—  \5bx 

,_    -1   ,         7a*rst:>      56«ra— I4ams       .     . 

13.  Reduce    --— — — r — to  its  lowest  terms. 

35a2ra  +  2 1  abm  +  56am 

,    _    ,  3Za3ry{+i8ax2y— 6ay  .    . 

14.  Reduce  -—  A    J?    '        1g   ,  q    J      to  its  lowest  terms. 

3a4i/*— 2 1  a^y*—  1  oa  ?/2— 9ay 

_    ,        112a&.x  —  48aca'  +  100aca?       .     . 

15.  Reduce  — — r — = r- to  its  lowest  terms. 

4abca  —  Sacd+52acx 

§  216.  By  this  principle,  we  may  often  simplify  answers 
to  questions  in  division.     That  is,  we  may  put  the  divisor 


Questions.    What  if  the  numerator  or  denominator  is  a  compound 
^tantity  ?     What  application  has  reduction  of  fractions  to  division? 


§217.]  FRACTIONS.  145 

underneath  the  dividend,  so  as  to  make  a  fraction,  and  then 
reduce  that  fraction  to  its  lowest  terms. 

EXAMPLES. 

16.  Divide  x2—2xy+xy2  by  4xy.        Ans.  X~~  y"*"^  . 

4y 

17.  Divide  I0xy—20x—5y  by  —5x.       2xv—4:X—v 

Ans. ^ 2. 

x 

18.  Divide  7abx— 56a*xy  +  l4ax3  by  28a*bx!ly. 

19.  Divide  8amy3+lGa3xy—24aby!i  by  48a2y»—72ay. 

20.  Divide  35o36c—14aa?+42o3  by  21a6— 28a5#+7a4#8. 

21.  Divide  32arty-|-16#32/2-f-8:ri/3  by  24a#+48a3x3. 

22.  Divide  54a3&3+45a3&3— 27a363  by  24a563+30a3&5. 

23.  Divide  48xy+\2axy— I6ax  by  4x*y*+§ax— 8xy 

24.  Divide  28(a— b+x)   by  4m(a— &+#).        Ans.  - 

m 

25.  Divide  12cc?(ra— rc)   by  14ac(m— n). 

26.  Divide  6a/*(r+/>)   by  24a(r+jo). 

27.  Divide  (a  -f&)  (m+ri)   by  (a— a;)  (m+n) 

28.  Divide  32a6c   by  32abcx—32aby. 

29.  Reduce  the  fractions  K^-\    tk^t*'>    "4^:1    n* 

28a*y      \2d?b*        hxy       5ab 


12ra3n  — 12mn3 


§  217.  It  must  be  remembered  that  when  all  the  factors  in 
the  numerator  are  contained  in  the  denominator,  the  answer 

will  contain   1   in  the  numerator.     Thus,   - — =— .      The 

Sax      x 

same  principle  will  also  apply  to  the  denominator. 

rp,         Aax      x 
Thus,  — —  =--  =  #. 
4a       1 

Question.     What,  if  in  reduction,  the  common  divisor  is  the 
as  one  of  the  terms  of  the  fraction  ? 
K  13 


146  ALGEBRA.  [EQ.  SEC.  6  &  7 


VI.    MULTIPLICATION    WHERE    ONE    FACTOR   IS    A   FRACTION. 

§  218.  This  is  done,  (as  shown  §  74,  75,)  by  multiplying 

the  whole  number  and  the  numerator  of  the  fraction  together, 

and  dividing  by  the  denominator. 

mi                3b  — x     6ab—2ax     a,      ab 
Thus,  2a  X  =  - J    7&  =  -r« 


EXAMPLES. 


1.  Multiply  ^  by  3a.  Ans.  —-^  =  xy 

2.  Multiply  - —  by  8a.  Ans.  — 

«„i-1n-  ab      .  .  ab 

3.  Multiply  3ax  into  —  — — .  Ans.  —  — 

1  'ZCIX  4 

,-«-,,,         .'  .      2by  iab*y—6bxy' 

4.  MulUply  2ab-3xy  by  g,         Ans.  -  ^--HI 

a&v  2a8#v — Saby* 

5.  Multiply  -- 2  by  2ax—3xy.        Ans.  ^ ^- 

C3?  C 

3x34-  a 

6.  Multiply  3am»— 4a?  by  . 

-    »«■  i  •  i     5as£ — 2m   .  "    . 

7.  Multiply  — by  am— 2a9. 

8.  Multiply  Wy+tafi  by  g^jy 

9.  Multiply sr-   by  2aa?-f  3a?s. 

Sax — ox 

10.  Multiply  I4abc—3cdx  by . 

r  J  5mx— Icm 


Questions.  How  is  multiplication  performed  when  one  factor  is  a 
fraction  ?  What  is  a  factor  ?  What  if  the  whole  number  is  a  factor 
in  the  denominator  of  the  fraction  1 


§  219.]  FRACTIONS.  147 

,,  .  .  ,     \0ax—6xy  ,      n  _ 

11.  Multiply  — — -   by  3mx— 2am 

12.  Multiply  -  by  b.  Ans.  -r-^a- 

ab  abc      ab 

13.  Multiply  —   by  c.  Ans.  — -  =-j- 

§  219.  In  the  last  example,  we  first  multiplied  the  nume- 
rator by  c,  and  then  divided  both  the  numerator  and  denomi- 
nator by  c.  Now,  multiplying  the  numerator  by  c  and  then 
dividing  it  by  c,  is  altogether  useless  ;  because  the  numerator 
is  left  as  it  was  at  first.  We  will  use  then  only  one  part  of 
the  operation  ;  that  is,  dividing  the  denominator  by  c.  And 
in  general,  when  the  multiplier  is  a  factor  in  the  denomina- 
tor, the  multiplication  is  performed  by  canceling  that  factor. 

EXAMPLES. 

,  i    „  ,.  ,     o,rnx  .  amx 

14.  Multiply by  r.  Ans.  . 

rs  s 

, ..    »•  i  .  i       o  i      obc  ,  abc 

15.  Multiply  x9  by  — — .  Ans.  — . 

x  y  xy 

16.  Multiply  ^  by  2m.  Ans.  29*. 

17.  Multiply  2d  by  |^. 

18.  Multiply   -—I —  by  2x. 

19.  Multiply  36*   by  a^±??. 

flWl8 

20-  ****  ui&=i§5*  by  to"- 

«,     n/r  t  •  ,     «      i  6am — 4a: 

21.  Multiply  3a   by 


3am  +  6ax  — 12a3 


~,    »,  ,  •  ,  abc  —  bx+cx 

22.  Multiply — -    ,       „ by  3aa?. 

VJ  9axy  +  l2a3x— Qaxc     J 


148  ALGEBRA.  [EQ.  SEC.  6  &  7. 

,  .  .  .         ,    ,  am-\-4ab — 2m 

23.  Muluply  iab  by  Sabm+Uab_4abx- 

24.  MuLtiply  - — ■ —    by  5a?— b. 

r  '  5a?-f  ay     J 


VII.    REDUCTION    OF    COMPLEX    FRACTIONS    TO    SIMPLE    ONES. 

§  220.  We  have  shown,  that  when  we  multiply  a  fraction 
by  its  denominator,  we  obtain  for  the  answer  the  same  quan- 
tity as  the  numerator.  We  have  also  shown  that  where  both 
terms  of  a  fraction  are  multiplied  by  the  same  quantity,  the 
value  of  the  fraction  is  not  altered.  By  these  two  principles, 
we  obtain  the  following  rule  for  reducing  a  complex  fraction 
to  a  simple  one. 

§  221.  Multiply  both  terms  of  the  fraction  by  the  deno- 
minator that  is  found  either  in  the  entire  numerator  or  de- 
nominator. If  the  fraction  is  still  complex,  multiply  the 
result  in  both  terms  by  the  remaining  denominator  that  is 
found  in  the  entire  term.     Thus, 

a  +-*  _  ca  +  b  ;        a  —  J      ca  —  b  ■     4ac  —  4b 
~1  cd     !        |T«  —  £  +  ca       3c -f4ac* 

EXAMPLES. 

1.  Reduce  ^-to  a  simple  fraction. 

*  ac 

Ans.  Multiplying  both  terms  by  c,      — r. 

a — —  .      ,    /.      .  ay  —  x 

2.  Reduce  *  to  a  simple  fraction.  Ans.     ° . 

ax  axy 


Questions.  How  may  a  complex  fraction  be  changed  to  a  simple 
fraction?  Explain  why.  How  may  fractions  be  transferred  from 
the  numerator  to  the  denominator ! 


§  222.]  FRACTIONS.  149 

X 

3.  Reduce  —   to  a  simple  fraction.  *v 

«2/+V  Ans. 


5ay+y—z 

x  —  —  4x—x      3 

4.  Reduce -  to  a  simDle  fraction.     Ans.  — =  -^* 

xy  4yx       4y 

5.  Reduce ^—t  to  a  simple  fraction.     Ans. y      - 

3a  —  -—  *ay   y-r*- 

x  +  - 

6.  Reduce  -  to  a  simple  fraction. 

Z -  x 

e  a?+-       xy  +  x  _cxy-\-cx 

z  —  l.       yz  —  \       cyz-y* 

4  +  -  .  24*  -f  Gx 

7.  Reduce  :  to  a  simple  fraction.        Ans. — . 

x  —  f  r  Gxz  —  bz 

8.  Reduce =J  to  a  simple  fraction.     Ans.  —    ,      '  • 

a  +  ^  acy  +  bxy 

3 +  4  .      t    „      .  3a?v  +  4v 

9.  Reduce =-  to  a  simple  fraction.         Ans.  — - -• 

4  —  -  4;ri/  —  5a? 

10.  Reduce  -  to  a  simple  fraction.      Ans.  = 3-. 

§  222.  It  sometimes  happens  that  we  wish  to  transfer  a 
fraction  from  a  numerator  to  the  denominator,  or  from  a  de- 
nominator to  the  numerator.     This  may  be  done  by  the  fore- 

2a 
going  principles.     For,  supposing  we  have  — ;  multiplying 

<*.  & 

On 

by  the  denominator  of  f-,  we  have  — .     Now,  if  we  divide 

IX 

this  fraction  by  2,  we  have  — .     Thus  we  see  that  such  a 

fraction  is  transferred,  without  altering  the  value  of  the  whole 
quantity,  if  we  take  care  to  invert  it  when  we  transfer  it. 


13* 


I5C  ALGEBRA.  [EQ.  SEC.  6  <fc  7 


EXERCISES  IN  EQUATIONS. 

§  223-  Generalize  the  questions  in  section  6th,  page  52. 

1.  In  an  orchard,  —  (\)  of  the  trees  bear  apples;  -  (|) 
m  v  '  n  V5/ 

of  them  bear  pears ;  -  (-fc)  of  them,  plums ;  and  a  (81)  bear 
cherries.     How  many  trees  are  there  in  the  orchard  ? 

ICT*  We  rarely  represent  unity  by  a  letter  ;  but  generally 
use  its  own  character,  1. 

Let    x  =  number  of  trees. 

—  =  apple-trees. 

x 

—  =  pear-trees. 


px        . 

' —  =  plum-trees. 


Then,  x=*  +  *+1E+a. 

m     n      r 

§  224.  As  the  equation  is  to  be  multiplied  by  m,  and  then 
by  n,  and  then  by  r ;  it  is  plain  we  may  multiply  it  at  once 
by  their  product,  mnr.  Of  course,  in  multiplying  a  fraction, 
its  denominator  will  not  appear  in  the  product.  §219. 
Multiplying  by  mnr,  mnrx  =  nrx-\-mrx-\-mnpx+mnra 
Transposing,  mnrx  —  nrx  —  mrx  —  mnpx  =  mnra 

Dividing  by  mnr  —  nr  —  mr  —  mnp, 

mnra 
a?  = 


mnr  —  nr  —  mr  —  mnp 
Substitute  figures  for  letters,  and  find  the  answer. 
4x5x11x81 


a?  = 


4x5x11  —5x11  -4x11  —4x5x2 

17820 

220  —  55  —  44—40  ' 

17820 
_-  =  220. 


§  224.]  FRACTIONS.  151 

2d  sum,  page  53.     In  a  certain  school,  —  of  the  boys  learn 

m  * 

P 
mathematics  ;  —  of  them  study  Latin  and  Greek ;  and  a  study 

grammar.     What  is  the  whole  number  of  scholars  ? 

mi  X  PX 

1  he  equation  is,  x  =  — f-- — \-a. 

m      n 

Multiplying  by  mn,  mux  =  nx-{-  mpx  +  mna. 

Transposing,  mnx  —  nx  —  mpx  =  mna. 

t\-t      u  mna 

Dividing  by  mn  —  n  —  mp,  x  = . 

r  mn  —  n  —  mp 

3.  The  pupil  must  go  through  with  the  whole  section  in 
the  same  manner.  And  as  there  are  several  instances  in 
which  the  same  statement  and  the  same  answer  will  agree 
with  two  or  more  sums,  the  pupil  must  tell  which  they  are, 
and  why  it  so  happens. 

4.  If  the  teacher  should  think  his  pupils  need  more  prac- 
tice, he  may  exercise  them  in  the  7th  section  in  the  same 
manner. 

The  first  question  may  be  stated  thus, 

x  +  a 

—  =c 

Multiplying  by  b,  x  +  a  =  bc 

Transposing,  x=bc  —  a. 

The  2d  will  have  the  following  equation. 

mx  -f-  ma      . 

=6 

n 

Multiplying  by  n,  mx  +  ma  =  nb 

nb  —  ma 


Transposing  and  dividing,  x 


m 
The  3d  and  5th  will  have  the  following  equation. 

nx  —  na      , 
=  o. 


The  4th  will  be,  a  —  x  =  — . 

m 


152  ALGEBRA.  [EQ.  SEC.  8. 


VIII.    DIVISION    OF    FRACTIONS. 


§  225.  We  have  shown  in  §  97,  that  we  divide  fractions 
by  dividing  the  numerator,  when  the  divisor  is  a  factor  in 
the  numerator  ;  but  if  the  divisor  is  not  a  factor  in  the  nu- 
merator, we  multiply  the  denominator  by  the  divisor. 


EXAMPLES. 

,        TV  Sa°      U  A  Sb 

1.  Divide  by  a.  Ans.  — . 

x       *  x 

M    _.  .,     Sax  ,      ,  Sax 

2.  Divide  -j—    by  o.  Ans.  -= — . 

be      *  b»c 

3.  Divide    ^ ' "~        by  3a:.  Ans.  ^r— 

4.  Divide  -2  by  —3a.  Ans.  —  -3L* 

z  Saz 

5.  Divide  95   by  7a3. 

6.  Divide.  ^  by  7. 

4z       J 

«,    t^-    j.  3ax  —  Gay   .      r, 

7.  Divide  — ^   by  2#y. 

4a;ry 

8.  Divide         ~^  by  3aw- 

a+y       J       u 

m    _.  ..    4a+27a&  t      - 

9.  Divide  — by  6ax. 

Zx 

-,n    r^-3     6abx—\6axy  .      n 

10.  Divide  — — — £   by  2a. 

2am-r3xy        J 

„.  Divide  146c3;+21ac3  by  7<ft 

4abx  J 


Question.    How  do  we  divide  fractions  by  whole  numbers  ? 


§  226.]  FRACTIONS.  153 

,«    t%-    j     10a5+53?5—  15ax  .      m  , 

12.  Divide  — 7  — !-— - — — —      by  5a5x*. 

a4+&x2—lQax       J 

,«    t..  -j     12a&  — 14c3?  .      rt  .. 

13.  Divide  — ■ ,  ,_,     by  2ac  —  4bx. 

10a.r  +  166c     ' 

14.  Divide by  4ca  —  cax. 

x  —  y 

15.  Divide  -±- by  12a3?— 9a3. 

16.  Divide    * "       ,*     by  a3-l. 

Ha3— 16     J 


IX.    FRACTIONS    OF    FRACTIONS. 

§  226-  It  was  shown,  §98,  that  a  fraction  is  multiplied 
by  a  fraction,  by  multiplying  the  numerators  together  for  a 
new  numerator,  and  the  denominators  together  for  a  new 

denominator.     Thus,  =-  X  -y  =  ■=-;. 
o     a       oa 

EXAMPLES. 

,    „  ,  .  ,     3«  L     4i  .         I2ab 

1.  Multiply  —   by  — .  Ans.  — — . 

037  C  OCX 

n    „  ,  .  ,     9ax  ,      2&3?  .         18C&3?3      6a?3 

2.  Multiply  —   by  — .  Ans.    ^  m  _ 

3.  Find  the  product  of  ^t^  into  — 2- .      Ans.  Xy  +y 

r  3a  x — y  ax — ay 

S3?  43? 

4.  Multiply  —  and  —  together. 

5.  Multiply  ~  into  -^. 

Sy  73? 

3a  4a 

6.  Multiply  ~r  into -. 

7.  Multiply  -  into  -^r". 


154  ALGEBRA.  [EQ.  SEC.  8. 

8.  Multiply  —  -  into  —  ^.  Ans.  ^. 

X  z  xz 

a    Sx   4y  ,  4a 

9.  Multiply  -,  — ,  -^  together.  Ans.  — . 

x     y     oZ  z 

.     ,  .         „  4ax   3a*y       .  2  . 

10.  What  is  the  product  of ,  -~y  and  -  ?     Ans.  12a:. 

r  y      2a  x 

0/»  2ac—~- be 

11.  What  is  the  product  of    ,         into  — —, —  ? 

„  ,  .  ,     2am3— 3a2m  ,         5am3 

13.  Multiply  '"f'+fy  by    **?-**  • 

r  #       4a:— 3y         J   5xy*+l0x*y 

§  227.  When  the  numerator  of  one  of  the  factors,  and  the 
denominator  of  the  other,  can  both  be  divided  by  the  same 
quantity,  it  is  best  to  perform  that  division  before  the  quanti- 
ties are  multiplied.  This  is  done  by  canceling  the  quantity 
that  is  common  to  the  numerator  of  one  and  denominator  of 

the  other.     Thus,  -rX —  = =  —  • 

cd    ay       cxy        cy 

,  '     .,  .  .  :     2x  ,      5a;9  .        a?8 

14.  Multiply  —  by  -— .  Ans.  — . 

2a:    Sab    5ac3 

15.  What  is  the  product  of  — x — X-rr-?      Ans«  15aa?. 

v  a       c       2b 

16.  Multiply  I*"*3!*'  by  10a6-2c.        1K 

*  J   10a36— 2ac     J  I5ac+37bc 

Ans.  ■ . 

a 

,„    ™  i.-  i     13a&— 13am  ,        17a  . 

17.  Multiply  r—; by  -. Ans.  13. 

r '  17a3  J   6  — m 

«8^_^2        q% ^3  qZ 

18.  What  is  the  product  of  — -j-x — -QX ?§185 

r  a+b      ax+x2    a — x 

a*—a*b 
Ans.  . 


§  229.]  FRACTIONS.  155 


EXERCISES  IN  EQUATIONS. 

§  228.  Generalize  the  questions  in  Equations,  section  8. 

1.  The  1st  sum  on  page  64  is  performed  as  follows ; 

Stating  the  question,  x  =  oats. 

mx 

—  =  barley. 

n  J 

1     „  mx      mx 

—  of  —  =  —  =  rye. 

p        n        np 

fTfV      TflX 

Forming  the  equation,       x-\ — -H =  a. 

§  229.  In  this  and  similar  equations,  we  may  apply  the 
principle  explained  in  §88.  For,  if  we  multiply  by  the 
greatest  denominator,  np,  it  is  plain  that  the  second  term 

will  be   — — ,  which  =  mpx. 

n  r 

Multiplying  by  np,  npx+mpx+mx  =  anp 

Dividing  by  np  4-  mp  -f  m,  x  = ■ — ~ — . 

&    J    r       r  np+mp+m 

x       x 

2.  The  equation  in  the  2d  sum  will  be =  a. 

n      mn 

Multiplying  by  mn,  mx — x  =  amn 

t\-  -j-      1.             i                                                amn 
Dividingbym — 1,  #  = -. 

3.  The  equation  in  the  3d  sum  will  be 

x    ,  x        x 

x= h«» 

m      n      mn 

4.  The  equation  of  the  4th  sum  will  be 

mx   ,  mx     m2x 

x= +  a. 

n         n         n* 

The  pupil  will  now  understand  how  to  perform  the  rest  of 
the  section. 

The  problems  under  ratio,  in  section  10,  page  78,  are  very 
good  examples  for  this  exercise. 


156  ALGEBRA.  [EQ.  SEC.  9 


X.    UNITING    FRACTIONS    OF    DIFFERENT    DENOMINATORS. 

§  230.  Before  fractional  terms  can  be  united,  they  must  be 
brought  to  a  common  denominator,  according  to  the  principle 
explained  in  §  85.  This  is  done,  as  we  have  shown  in  §  104, 
105,  by  multiplying  each  numerator  by  all  the  denominators 
except  its  own,  for  new  numerators;  and  all  the  denomina- 
tors together  for  a  new  denominator, 

EXAMPLES. 

I.  Unite  the  following  terms, --{-  -—-. 

X  *xU        ox 

8ab*x — abxB +4abmx         Sab*— ax2 + 4am 


4b»x*  4bx 

2.  Add and together.    §  178.      Ans.    

z-f-l  x—l      6  x*—  1 

o    o  u*      .  Sa  r        4a  a        S2ac—2lab 

3.  Subtract  — -  from  —p.  Ans. 


8c  76  566c 


4.  Unite 

x+y 

y 

z 

5.  Unite  X~V- 

x+y 

Ans.  ^-^-y 


xz+yz 

ax — Say 

Ans.  — „  „  J 


2a  3a  6a3 

§231.  If,  in  such  cases,  there  is  a  quantity  that  is  no 
fractional,  we  multiply  it  by  all  the  denominators ;  and  then 
putting  the  common  denominator  under  that  product,  unite  i 
with  the  other  fractions. 

n    tt  •         ■  °      l  ca        i  !        ca      b      ac+b 

6.  Unite  a-\ — .     Ans.  a  =  —  ;  and  then, j — == 

c  c  c       c  c 

Questions.  How  can  fractional  terms  be  united  ?  How  can  we 
bring  them  to  a  common  denominator  ?  Supposing  there  are  quan- 
tities connected  with  them  that  are  not  fractional?  How  do  we 
reduce  fractions  to  the  least  common  denominator  ? 


§232.]  FRACTIONS.  157 

7.  Unite  x  -\ ■ — .  Ans.  -^ — ■ — 

y  y 

■m    TT  .-  a-\-b  ac—a—b 

8.  Unite  a .  Ans.  . 

c  c 

§  232.  It  is  generally  best  to  reduce  the  fractions  to  the 
least  common  denominator.  This  is  the  least  common  mul- 
tiple of  all  the  denominators ;  and  is  found,  when  the  same 
quantity  is  in  two  or  more  denominators,  by  taking  that  quan- 
tity but  once  in  the  required  product.     Thus, 

n    TT  .        ar         2c         3cd 

9.  Unite  — — +  — — — _ 

5om2      5bn      2msn 

Ans.  The  common  denominator  is  5x2x&X  m3xn- 
=  I0bm3n.  To  obtain  this,  the  first  denominator  must  be 
multiplied  by  2mn;  and  therefore  the  .numerator  must  be 
multiplied  by  the  same,  and  becomes  2amnr.  The  second 
denominator  must  be  multiplied  by  2m3;  and  therefore  its 
numerator  must  be  multiplied  by  the  same,  and  becomes  4cm3. 
The  third  denominator  must  be  multiplied  by  5b  ;  and  there- 
fore its  numerator  must  be  multiplied  by  the  same,  and  be- 

2amnr-\-4cm3—  \5bcd 


comes  \5bcd.     The  answer  is 


.  ';—  :      3m95        2ar 
10.  Unite  -— -f- 


I0bm3n 


n5rs      3mn3r 

15a        be         2m8r 

11.  Unite—  +  _+^-. 

,«     AJJ  13a36— 2c       .    7ab+8c 

12.  Add  — r —    and  n.     7„  _ 

4ab  2b  +  16ab 

cde  3c d 


13.  Unite 


2lbm3x2      lb3m2x 


WET  To  know  what  any  denominator  must  be  multiplied 
by,  to  make  the  least  common  denominator,  divide  all  the 
other  denominators  by  such  factors  of  this  as  will  divide 
them,  and  then  multiply  this  by  the  several  quotients. 


158 


ALGEBRA. 

2abd—3m*c 


[eq.  sec.  9, 


.a    ^        Wad    i 

14.  From-rrj-.  take  — 

t_    _  a+x  a—x 

15.  From  —. r  take 


a(a— x)  "*~"°  a(axx) 
a-f-6       a — b 


,«    4JJfl+^    a—b        A        2a9+2&9        ol     469 
16.  Add  — 5-r  to  — — r.      Ans.  __-!-— --  =  2+— — r- 
a— 6       a  +  6  a9— o9  a9— 69 


17.  From  — :  take 


a»—z* 

•8 


Ans. 


3a,r— a8— z9 
a3— *a 


a-f-z 
'a&       ,     6         .        as+a68+68 

18-  Umte  (^"(W'+S+S'   Ans-  "ISP^ST 

■     ,T  .         .a  4         .        7bx+7a—7aby—4b 

19.  Unite  *+-r  —  ay  —  -.      Ans.   -^ — ^ 

v      5«3+6        ,4a9+26      A       37a9-fll6 

20.  Add  together        ,     ,  and  — -r Ans. ^ 


Ans. 


127#+17 


Ans. 


28 

4a?fl— llx— 5 
5a?+5       ' 

2a?9 


5*4-1  f         21x+3 

21.  Subtract  -  from  — - — . 

3z+l  -        4x 

22.  Subtract  — —  from  — . 

x+l  o 

23.  Add  together^,  and  -J-. 

,    _  ,  2a?  — 3  r        4a?+2 

24.  Subtract  — - from  — - — . 

3x  3 

25.  Subtract  ^—7  from  ^t_. 

a+o  a — 0 

rtrt    _         3a+26       ,  5bd—2a  —  3d 

26.  From  ,  subtract  — -, . 

c  4cd 

_  *      '2     •  ;,  69c  — 5aft9c+a8 

27.  From  c+2o6  — 3ac,  subtract  j- — ^ . 

exercises  in  equations. 

§  233.  Generalize  the  questions  in  Equations,  section  9, 
page  70. 


Ans. 
Ans. 
Ans. 


a:9— 9 

4a?9 +3 
3x 

4ab 


§234.]  FRACTIONS.  159 


XI.    DIVISION    BY    FRACTIONS. 

§  234.  Suppose  we  wish  to  know  how  many  times  ^  is 
contained  in  f-.  We  would  divide  in  the  same  manner  that 
we  follow  in  dividing  6  pieces  by  3  pieces;  and  say,  ^  is  con- 
tained in  f,  two  times.  In  the  same  manner,  ¥4r  is  contained 
in  if, Jive  times. 

The  principle  is  general  that  when  the  divisor  and  the 
dividend  have  a  common  denominator,  the  division  is  per- 
formed by  dividing  the  numerator  of  the  dividend  by  the 
numerator  of  the  divisor. 

,„,        a      c       a       a*      a2      a3 
Thus,  T -j-T  =  -;     —  -i- _  =  _  =  a. 
bo       c        x       x       a* 

EXAMPLES. 

|    _.  .,     4a  .      2b  Aa      2a 

1.  Divide  —  by  — .  Ans.  ^-r  =  -i— 

x      J    x  26       b 

n    _.  .,     Sab  ,      4bx  3ab      3a 

2.  Divide  — r-   by  — r.  Ans.  -z—  =  — — 

cd      *    cd  4bx      4x 

'    ^.  .„    2x  ,      3xy  2 

3.  Divide  —   by  — £.  Ans.  — . 

ab     J    ab  3y 

.    —  .,    3a  — b  .      2a»  A        3a  —  b 

4.  Divide  —   by  — j-.  Ans.  ■  * 

aft         y    a6  2a2 

.    _.  .,    7r#+a9   •■     46-faa?  .        7rrr-faa 

5.  Divide      ,    '        by     . '       .  Ans.  — =— 

bast        f      bast  4b-\-ax 

„    t^.  • ,    «     ,      2a6 

6.  Divide  8a  by 

Explanation.     By  §  231,  8a  =  -^.     Then  ~22?-|— 

^  '  »  xy  xy       xy 

Saxy      4xy 
"  2a7T  *"  T~* 


1?  A«.     *S 

2/ 


7.  Divide  3a  by  — .  Ans. 


160  ALGEBRA.  [EQ.  SEC.  11. 

8.  Divide  ab  by  -3.  Ans.  . 

a  c 

_    _.  .       4a*x         2x*  2a* 

9.  Divide  by  — .  Ans.  — . 

y  y  # 

in     TV    'A      8mnr    u       6flm*  a  4nr 

10.  Divide  by  .  Ans.  - — . 

ax        *     ax  Sam 

ii  tv  -a  4av*x  u    l0ax*  %v* 

11.  Divide  -f—   by  —r — .  Ans.  --^-. 

be        J      be  5x 

i«    tv    a    a+6c  u     4a9— «&  A  a+oc 

12.  Divide  by  .  Ans. 


ax       *        ax  4a*— ab 

lO      TV      A  am*        U  QX  "** 

13.  Dmde  g-^  by  ^--.  Am.  -. 

U.  Divide  ^=^  by  % 

ab         *     b 

Explanation.     In  this  example,  the  dividend  and  divisor 

has  not  a  common  denominator.     Our  first  object,  then,  is  to 

bring  them  to  a  common  denominator.   This  is  done  by  §230, 

ax — y      xy       bax  —  by      abxy 

ab      ~T  ""       ab*       ~  ab*  ' 

.        abx  —  by       ax  —  y 

Ans.    — r ^  = £. 

abxy  axy 

15.  Divide  —   by  -^. 
c        .a 

~  abx      dy  __  a*bx      cdy  __  a*bx 

v  e     '   a  '      ac     '   ac        cdy 

§235.  It  will  be  seen  that  in  sums  of  this  kind,  after  we 
have  brought  the  terms  to  a  common  denominator,  the  division 
is  performed  by  putting  the  numerator  of  the  dividend  for  the 
numerator  of  the  answer,  and  the  numerator  of  the  divisor  for 
the  denominator  of  the  answer,  and  make  no  use  at  all  of  the 
denominators.  Let  us  see  then  how  we  obtain  these  two 
terms.     We  multiply  the  numerator  of  the  dividend  by  the 


§  238.]  FRACTIONS.  161 

denominator  of  the  divisor ;  and  this  becomes  the  numerator 
of  the  answer.  And  we  multiply  the  numerator  of  the  divisor 
by  the  denominator  of  the  dividend ;  and  this  becomes  the 
denominator  of  the  answer.  By  looking  at  the  last  two 
sums,  it  will  be  seen  that  this  is  the  true  operation. 

§  236-  Hence  we  obtain  the  general  rule  for  dividing  by 
a  fraction.  Multiply  the  numerator  of  the  dividend  by  the 
denominator  of  the  divisor,  for  a  new  numerator  ;  and  mul- 
tiply the  denominator  of  the  dividend  by  the  numerator  of 
the  divisor  for  a  new  denominator. 

§237-   When   the   dividend  is  a  whole  number,  it  is 

changed  into  a  fraction  by  putting  1  under  it  for  a  denom- 

x  2a 

inator.     Thus,  x  —  -  ;         2a  =  — . 

EXAMPLES. 


16. 

3x         ax 

Divide  — -  by  — T. 
4      '    26 

66*       36 

Ans.  - —  =  —• 
4ax      2a 

17. 

Divide  "  +  X  by  -a-+y 
a — y         a-\-2x 

a2  +  Zax+2x* 
Ans.  —  - 

18. 

Divide  x2 — 2ax-\-a3  by 

1 
x — a' 

Ans.  x 

3  —  3#Ba-f  3a*x  —  a*. 

§  238.  If  both  of  the  numerators,  or  both  of  the  denomi- 
nators have  the  same  factor,  that  factor  may  be  canceled  be- 
fore the  operation  is  performed. 

^        ax     ay      x      y         xt 

Thus, 1.—  =  — 7--  -  ■■  — • 

rs      rt        s      t         sy 

,  _    _..  ..     a   .      3a  5 

19.  Divide  -    by  — .  Ans.  —  • 

Questions.  What  are  the  rules  for  dividing  by  fractions?  Explain 
the  rule  for  fractions  that  are  not  of  the  same  denominator.  Sup- 
posing the  dividend  is  a  whole  number'?  In  what  cases  can  the  work 
be  shortened  1 

L  !4* 


52                                                       ALGEBRA.  [EQ.  SEC.  11. 

«rt     ™-      J      X— l     u       X+l  A            4X~ 4 

20.  Divide  —    by  _.  Ans.  -j—. 

21.  Divide  by  —r.  Ans.    — . 

y            46               #,  5ay 

tai  . ,    5a&  .          4a  .             5% 

22.  Divide  — -    by Ans.  —  -£. 

4                y  16 

~.  . ,       a      .      3x  4ay 

23.  Divide by  — .  Ans. f~- . 

a+1      *   4y  3a#-f3a? 

^.  .,                 .         3a  x*+axa—xy—axy 

24.  Divide  a?-r-a;r  by  .     Ans.   — ' -—* *. 

y    x— y  3a 

25.  Divide  -JL.    by  ^.  Ans.  -. 

y— i       3  y— * 

*     _.  .,                   ,      Ay— ay  16a9— 4a3y 

26.  Divide  4a— ay  by  -^- — ^.  Ans.   — «. 

a               4a  Ay  — ay 

27.  Divide  by    ■- -.   See  §  177,185.  Ans.  ;r+y. 

x — y         x  *— y*  y 

na    r»-    j    ai  —  xi  .        a+a? 

28.  Divide  — -=-   by  .    ;  ,.  . 

a+b       J   (a+b)* 


XII.  GENERAL  THEORY  OF  EQUATIONS  WITH  TWO  UNKNOWN 
QUANTITIES. 

§239.  Every  equation  of  the  first  degree  with  two  un- 
known quantities  may  be  reduced  so  as  to  be  represented  by 

ax+by  =  c, 
a  designating  the  algebraic,  sum  by  which  one  unknown  quan- 
tity is  multiplied;  and  b  the  sum  by  which  the  other  unknown 
quantity  is  multiplied  ;    and  c  the  algebraic  sum  of  all  the 
known  quantities. 

For  example,  the  equation  mr+nx-j-py-\-qy — ry=  l-\-kt 
iwv  be  changed  into  v  (m+n)x  -t-(p  +  q—r)y  =(/  +  &), 
where  (*n4-n)  =  a ;     (p  +  q— r)  =  b  ;    and  {l+k)  —  c. 


§  242.]  FRACTIONS.  163 

§  240*  If  there  are  two  equations  of  this  kind,  there  will 
be  two  sets  of  co-efficients  and  whole  numbers.  But  for  our 
present  purpose  it  will  be  best  to  represent  both  sets  by  the 
same  letters ;  only  accompanying  the  second  set  with  the 
accent,  to  intimate  that  they  stand  for  different  quantities  from 
the  first  set.     The  two  equations  will  then  become 

ax  -\-by  =c 

a'x-\-b'y  =  c\ 

§  241.  In  order  to  make  the  terms  of  y  identical,  by  the 
rule  for  common  denominators,  §231,  we  will  multiply  the 
first  equation  by  b',  and  the  second  by  b,  when  we  shall  have 

ab'x-\-bb'y  =  b'c 
a'bx+bb'y  =  be' 


Subtracting  one  from  the  other,  (ab' — a'b)x  =  b'c — be' 

tv  li-  b'c- be' 

DlvldmS'  *rac35 

-^  .       ,  ,  .  .     ,  cb' — be' 

Or,  putting  the  accented  quantities  last,  x  = 


ab' — ba' 

By  the  same  means  we  shall  find     y  —  — =- — #-, 
J  v      ab'—ba' 

§  242.  By  the  use  of  accents  in  the  notation  of  co-effi- 
cients, the  pupil  may  readily  see  the  law  by  which  these  two 
formulae  may  be  found  without  performing  the  whole  opera- 
tion of  eliminating. 

The  common  denominator  is  found  by  forming  the  two 
arrangements  of  the  co-efficients  ab  and  ba,  with  the  sign  — 
betiveen  them;  and  then  accenting  the  last  letter  in  each 
term.     Thus,  ab'  —  ba'. 

To  obtain  the  numerator  for  answering  each  unknown 
quantity,  take  away  from  the  denominator  the  letter  which 
designates  the  co-efficient  of  that  unknown  quantity;  and 
put  in  its  place  the  letter  which  designates  the  known  quan- 
tity, and  then  accent  the  last  letter  in  each  term. 

Thus,  first  we  have  for  the  denominator,  ab — ba.  For  the 
answer  to  x,  we  will  take  away  the  a's  and  put  in  c's  ;  and 


164  ALGEBRA.  [EQ.  SEC.  11 

we  shall  have  cb — be,  which,  when  accented,  will  be  cb' — be', 
as  above. 

For  the  answer  to  y,  we  take  away  the  b's  and  put  in  c's ; 
and  we  shall  have  ac — ca,  which,  when  accented,  will  be 
ac' — ca'. 

§  243.  It  must  be  understood  that  a,  6,  and  c,  stand  for 
any  algebraic  quantities,  whether  positive  or  negative;  and 
whenever  those  quantities  are  replaced  for  a,  b,  and  c,  their 
own  signs  must  accompany  them  in  connection  with  the 
signs  of  the  above  formulas.  Thus,  let  there  be  two  equations : 
x  —  2y  =  8 
—  Sx  +  4y  =  —  58 
then  we  shall  have  a  —  1 ,  b  =  —  2;  c  =  8,  a'  =  —  3,  6'=4, 
c'  =  —  58 ;  and  by  substitution  in  the  formulas, 

_cb'  —  bc'     _'  —  8x4  —  (—  58x   -2)         32  -116 
X-ab'—ba'~~       l^4-.(-    8x-8)""'    4-6 

_acj-ca'    _-58xl-(     8x—  3)        —  58—  (—  24) 
y—ah'—ba'~         4x1—  (— 2x—  3)"  4  —  6 

Question  1.  Change  the  above  formula  for  the  denomi- 
nator common  to  their  two  values  into  a  rule. 

Question  2.  Change  the  formula  for  the  numerator  in  the 
value  of  x,  into  a  rule. 

Question  3.  Change  the  formula  for  the  numerator  in  the 
value  of  y,  into  a  rule. 

Examples.  The  pupil  may  perform  a  few  sums  in  section 
11,  page  88,  by  these  rules. 


§  248.]  INVOLUTION    AND    POWERS.  165 

xin. 

INVOLUTION  AND  POWERS. 

§  244.  We  have  already  shown  that  when  a  quantity  has 
been  multiplied  into  itself,  the  product  is  called  the  second 
power;  and  that  when  it  is  taken  three  times  as  a  factor,  the 
product  is  called  the  third  power  .'*  In  the  same  manner,  any 
number  of  times  with  which  a  quantity  is  taken  as  a  factor, 
will  give  its  name  to  the  product.     §  163,  164. 

§  245.  In  other  words,  the  exponent  of  any  quantity  repre- 
sents its  name  as  a  power.  Thus,  a  or  a1  is  the  first  power 
of  a ;  a2  the  second  power  of  a;  a*  the  eighth  power  of  a; 
an  the  nib.  power  of  a ;  2y\  ,  or  2?/8,  or  (2y)3,  is  the  third 
power  of  2y. 

§  246.  When  the  first  power  of  any  quantity  is  compared 
with  the  higher  powers,  it  is  called  the  root  of  those  higher 
powers ;  because  it  is  from  that  quantity  that  they  may  be 
said  to  grow.     Thus,  x  is  called  the  root  of  x2  and  of  #3,  &c. 

§  247.  Involution  is  finding  any  power  of  a  quantity.  // 
is  performed  by  multiplying  the  quantity  into  itself  till  it 
is  taken  as  a  factor  as  many  times  as  there  are  units  in  the 
exponent  of  the  required  power. 

§  248.  When  a  quantity  is  represented  by  a  single  letter, 
the  multiplication  is  performed  by  simply  annexing  the  re- 
quired exponent,  by  §  162.     But,  if  the  quantity  consists  of 


Questions.  What  rule  have  we  for  naming  powers?  How  is  the 
name  represented  on  paper]  What  is  called  the  root  of  powers? 
Why?  What  is  involution?  How  is  it  performed?  How  is  in- 
volution performed  on  single  letters? 

*  Sometimes,  for  the  sake  of  conciseness,  the  second  power  of  a 
quantity  is  called  the  square  of  that  quantity ;  and  the  third  power 
is  called  the  cube. 


166  ALGEBRA.  [eQ.  SEC.  15. 

two  or  more  factors,  the  exponent  must  be  annexed  to  each 
of  them  ;  and  if  there  is  a  numeral  factor,  the  multiplication 
must  be  actually  performed.  Thus,  the  second  power  of  ax 
is  found  to  be  a*x2 ;  the  third  power  of  2y  is  23y3,  or  Sy3 ; 
the  fourth  power  of  3ab  is  3*a4b*,  or  81a464. 

EXAMPLES. 

1.  What  is  equal  to  (2ab)5  ?  Ans.  25a5b5  =  32a56* 

2.  What  is  equal  to  (2ax)2  ?  Ans.  4aa#2. 

3.  What  is  equal  to  [aby)*1  Ans.  a3b3y*. 

4.  What  is  equal  to  (5mnx)*l  Ans.  625m4nAx*. 

§  249.  Hence  we  learn,  that  any  power  of  the  product 
of  several  factors,  is  equal  to  the  product  of  their  powers. 
Thus,  the  fourth  power  of  ab  is  (ab)*>  or  (a4b*).  The  third 
power  of  16  =  4096;  or  it  equals  (2x8)3  which  equals  23x8s 
which  =8x512  -  =  4096. 

§  250-  From  the  preceding  remarks,  we  derive  a  principle 
that  is  of  much  importance  in  mathematics.  It  is  the  follow- 
ing :  (2a)2  =  22a9,  or  4a3.  That  is,  four  times  the  second 
power  of  any  quantity  is  the  same  as  the  second  power  of 
twice  that  quantity;  as,  23(x+|)3  =  4(;r+§)3. 

§251.  We  see  that  when  an  exponent  is  to  affect  only 
one  letter,  it  is  annexed  to  that  letter  alone ;  but  when  it  is  to 
affect  a  quantity  which  is  represented  by  more  than  one  letter, 
that  quantity  must  first  be  enclosed  by  a  vinculum,  or  paren- 
thesis, and  then  the  exponent  is  annexed  to  that.  Thus,  the 
second  power  of  a  +  b  must  be  represented  by  either  (a-f  &)3> 
or  a-f-6|3,  or  a  +  b  .  The  parenthetical  form  is  generally  the 
best. 

Questions,  How  is  involution  performed  if  there  are  more  letters 
than  one?  What  if  there  is  a  numeral  factor?  What  does  the  rule, 
together  with  the  representation  of  the  power  show?  What  im- 
portant principle  is  derived  from  this  ?  Supposing  an  exponent  is  to 
effect  a  quantity  of  more  than  one  letter? 


§  252.]  INVOLUTION    AND    POWERS.  167 

§  252.  In  involving  compound  quantities,  it  is  found  best, 
for  most  purposes,  to  give  them  simply  the  proper  exponent. 
Thus,  the  fifth  power  of  2a— x  =  (2a — x)5.  But  there  are 
some  cases  in  which  it  is  necessary  to  perform  the  multipli- 
cation in  its  extent.  And  that  operation  is  called  expanding 
or  developing  the  value  of  the  expression. 

5.  Thus  we  expand  or  develope  (2a— x)5,  as  follows : 
2a— x 
2a — x 
4az—2ax 

—2dx  +  x* 
4a'i—4ax+x2  =  (2a— x)*. 
2a  —x 


8a3  —  8a2x  +  2ax* 

— 4a-x-{-4ax* — x% 
8a3—  i2a*x+Qax^-- x3  =  (2a— x)\ 
2a  — x 


16a4— 24a3x+  I2a2x2—2ax3 

— 8a*  x  +  12a'.r2— 6ax3+x* 
r6a4— 32a3x  -f  24a^a— 8aa?3  -f-  x*  =  (2a- 
2a  —x 


32a5—64a*x  -f  48a3#2—  1 6a  V  +  2ax* 

—  1 6a4x  +  32a3.r2— 24a2,  r5  -f  8ax4— x5 
32a5— 80aix  +  80a'x»—  40a2rJ+10a;r4— x5  =  (2a— a:)5 

EXAMPLES. 

6.  Expand  the  binomial  (a-|-2:r)4. 

Ans.  a4-f8a,#+24a2:r2+32ar,+  in;r4. 

7.  Expand  the  binomial  (2x—3y)s. 

Ans.  8x3—36x»y+54xy*r-21y3. 

8.  Expand  the  binomial  (3— x)4. 

Ans.  81— 108»+54x2— 12a?3+^. 


Questions.     What  are  we  said  to  do  in  performing  the  mulliplico 
Hon  in  involution?     Do  we  always  do  that  1 


168  ALGEBRA.  [EQ.  SEC.    15. 

9.  Expand  the  trinomial  (a+b — c)9. 

Ans.  a3+2a&— 2ac+b*— 2&c+ca 

10.  Expand  the  trinomial  (x9— 2a? +  l)3. 

Ans.  a?9— 6a?5  +  15a?4— 20a?3 +  1 5a?9— 6a* + 1. 

0C?*  If  the  binomial  has  but  one  letter  in  each  term,  the 
square  is  immediately  known,  from  §  179  and  §  180. 

§  253.  The  powers  of  a  fraction  are  found  by  raising  both 
numerator  and  denominator  to  the  power  required. 

Thus,  (•  y - g  of  f  of  t  -  2v ;  ay  -  3. 


EXAMPLES. 


11, 


What  is  the  sixth  power  of  j-  Ans. 


3a?  27a?* 

12.  What  is  the  third  power  of  —  ?  Ans.  —— 3. 

if  if 

13.  What  is  the  sixth  power  of  -^?  Ans.  — -£-. 

r  x  a?8 

14.  What  is  the  fourth  power  of  -^?  Ans.  n  fA 

r  5a    .  625a* 

§254.  It  was  shown,  §167,  that  a*xa*  =  a*;  and  that 
flsxa3xa3xa3  =  a,a.  In  each  of  these  cases,  the  exponent 
was  added  as  many  times  as  the  quantity  was  to  be  taken  as 
a  factor;  by  which  we  see  that  when  a  quantity  has  already 
an  exponent,  it  is  raised  to  any  power  by  multiplying  the 
exponent  by  the  exponent  of  the  required  power.  Thus,  the 
"ourth  power  of  a3  is  a3X4  =  a13. 

§  255.  As  to  the  signs  of  the  powers,  by  the  principle  of 
multiplication,  §  172,  if  the  root  is  plus  (  +  ),  all  the  powers 
are  plus;  but  if  the  root  is  minus  (—-),  all  the  even  powers 
are  plus,  and  all  the  odd  powers  are  minus.  The  second 
power  of  —a  is  —ax  —a,  which  is  -fa3 ;  the  third  power  is 

Questions.  How  do  we  find  the  powers  of  fractions  1  How  do 
we  involve  a  quantity  that  has  an  exponent?  What  is  the  rule  for 
signs  of  the  different  powers  * 


§  256.]                          INVOLUTION    AND    POWERS  169 

-\-aax — a,  which  is  — a3;  the  fourth  power  is  — a*X — «» 
which  is  -fa4 ;  the  fifth  power  is  —a5. 

EXAMPLES. 

15.  What  is  the  third  power  of  6V?  Ans.  b«x9. 

16.  What  is  the  fourth  power  of  a*y*xt  Ans.  a™y8x4. 

17.  What  is  the  fifth  power  of  bc3x»1  Ans.  b*ci5xi0. 

18.  What  is  the  third  power  of  ab*x2y1  Ans.  asb6z6y3. 

19.  What  is  the  third  power  of  —3x*y3 ?  Ans.  —27x6y9. 

„„      .     ,                        ,.      3a*x\  2187a14#31 

20.  What  is  the  7th  power  of -— ?    Ans. y  » 

3z  27z3 

22.  What  is  the  nth  power  of  a8  ?  Ans.  a3n. 

2xr2  Sx^r9 

23.  What  is  the  third  power  of  — —  ?  Ans.  — — — . 
r                 3y  27t/3 

24.  What  is  the  nth  power  of  — -  ?  Ans 


ax 
21.  What  is  the  third  power  of  —  — —  ?        Ans. 


ay1"  arymn 

w,      .     -.  i  n  —  a3x(d+m) , 

25.  What  is  the  second  power  of  — ; — ^-r- — '-  ? 

r  (x+iy 

(x-f  l)6 

26.  What  is  the  third  power  of —fa?3?/8?    Ans.  —  &xBy9. 

§  256.  If  we  divide  any  power  by  its  root,  we  obtain  the 

a5 
power  next  below.     Thus,  as-r-a  =  a*,  or—  =  a4.     Again, 

a*  a3  a3 

a4-r-a  =  a8,  or  —  =  a8;  —  ==a8;  —  =  a.     Let  us  proceed; 
a  a  a 

—  =.  1 ;  but  if  we  diminish  the  exponent,  as  we  did  in  the 

preceding  divisions,  we  have  -  =  a0.  Hence  we  learn,  that 
a0  is  equal  to  1,  however  great  or  small  the  value  of  a 
may  be. 

Questions.     What  power  of  any  quantity  is  equal  to  11     How  do 
we  learn  this  * 

15 


170  ALGEBRA.  [EQ.  SEC.  15. 

This  may  be  shown  in  numbers.  49  is  the  second  power 
of  7.  Now,  49-r-7  =  7,  theirs*  power;  and  7-7-7  =  1,  the 
no  power. 

§  257.  But  we  can  proceed  still  farther  with  our  division. 

l-r-a  =  -;    — 7-a  =  — :    — -a  =  —  ;  <fcc.    Or,  taking  a'\  in- 
a      a  a3      a9  a3  5 

stead  of  1,  and  dividing  by  subtracting  the  exponent  of  the  di- 
visor, §  190,  a°-7-a  =  a0_1  or  a'1 ;    a~1-7-a  =  cr1~1  or  a-3; 


§  258.  Thus,  by  pursuing  two  different  methods  of  di- 
viding, we  obtain  two  different  sets  of  expressions,  both  of 
which  are  equal  to  one  another.  We  will  arrange  them  one 
under  the  other,  so  that  any  one  in  the  upper  line  shall  be 
equal  to  the  corresponding  one  in  the  lower  line. 

1         11        1  1   ^ 


I 


a*     a3     a3     a1     I  -       ~i       — ;      -r       — 

a       a3       a3      a*       ac 


a5     a*     a3     a9     a1     a0         arx     a"3     or3     a*     a-5) 
Here,  the  quantities  on  the  right  of  1  in  the  upper  line,  and 
on  the  right  of  a°  in  the  lower  line,  are  called  the  reciprocal 
powers  of  a.     They  may  be  read  the^r^  reciprocal  power, 
the  second  reciprocal  power,  &c. 

§  259.  By  this  method  of  notation,  we  may  remove  the 
denominator  of  any  fraction.  For,  we  have  only  to  put  it  by 
the  side  of  the  numerator,  and  give  it  the  negative  exponent. 

IJhus,  —  —ax— ,-=flxra-  =  ar2.     So  -  =  -•■flX-r« 
▼^      x*  x*  x       xl  xl 

ssrfl.r'1 ;  where  we  see  that  if  the  denominator  has  not  any 

written  exponent,  1  is  understood,  and  must  be  written  when 

it  becomes  negative,  in  order  to  distinguish  the  quantity  from 

a  positive  one. 

Question.  Show  the  principle  by  numbers.  In  how  many  methods 
of  notation  can  we  continue  to  divide  by  the  root  1  What  are  the 
corresponding  quantities  below  1?  What  are  they  called1?  What 
are  we  enabled  to  do  by  these  two  methods  of  notation  1     How  1 


§261.]  INVOLUTION    AND    POWERS.  171 

EXAMPLES. 

Remove  the  denominators  from  the  following  fractions 

a 


27.  —  =  abn-ixr1. 
nx 

clx2 

28.  r— -  =  axty-ty-*. 
by*  * 

be 

29.  .    .  .  -o  bcar*m-*nr\ 


30.   r  —  a(a—b)-h 

a — b         v        ' 

32.  J£L-  =  m(2— a)"1. 
2 — a  v         ' 


§  260.  It  is  very  evident  that  these  quantities  with  nega- 
tive exponents  can  be  returned  to  the  denominator  without 
affecting  their  values.  And  on  the  same  principle,  any  re- 
ciprocal  power  in  the  numerator  may  be  removed  to  the 

ax~  2 
denominator.    For,  suppose  we  have  the  quantity  - — .    We 

know  that  it  is  equivalent  to  —  xar*  which  =  —  x  —  which 

by  by     x2 

__    a         ,  ax~2         a 
~~  byx2'  by    .- ■  byx2' 

T      l_  1  "c  ,  x 

In  the  same  manner,  ab~xc  ==  -p  '*   and  a~2x\r*  =  — * . 

b    .  u          a*y 

§261.  Powers  with  negative  exponents  are  involved  in 

the  same  manner  as  if  they  were  positive.     Thus,  the  second 

power  of  a~2  is  a-2*3  or  ar* ;  because  the  second  power  of 

—  is  —  =  ar4.     The  third  power  of  b~2  is  6~8,  &c. 
a2       a*  * 


EXAMPLES. 


K 


a8 

33.  What  is  the  fourth  power  of  cPb"1  ?     Ans.  a9b~*  =  j-^ 

x° 

34.  What  is  the  third  power  of  x*y~2  ?     Ans.  x9y-°  =  —  • 

if 

bmn 

35.  What  is  the  nth  power  of  bmx~m  ?  Ans.  — ji 

x 

Question.     What  if  there  are  reciprocal  powers  in  the  numerator? 
How  are  reciprocal  powers  involved  ] 


172  ALGEBRA.  [EQ.  SEC.  15. 

XIV. 

evolution: 

§  262.  Whenever  we  meet  with  a  quantity  that  is  a  power, 
we  know  that  there  must  be  some  other  quantity  which  is  the 
root  of  that  power.     See  §244,  246. 

The  operation  of  finding  the  root  of  any  power  is  called 
Evolution.  But  as  for  our  purposes,  it  will  be  best  to  sub- 
divide this  part  of  algebra,  we  shall  at  present  treat  of  so 
much  of  Evolution  as  relates  to  finding  the  xoot  by  mere  in- 
spection; and  leave  that  part  of  it  which  is  generally  called 
Extraction  of  Roots  for  a  future  chapter. 

§  263.  When  a  quantity  is  reduced  from  its  second  power 
to  its  root,  we  say  we  have  found  its  second  root ;  when  the 
reduction  is  from  the  third  power  to  the  root,  we  say  we  have 
found  the  third  root,  &c. 

§  264.  Definition. — The  root  of  any  quantity  is  a  fac- 
tor which  being  multiplied  into  itself  a  certain  number  of 
times  will  produce  that  quantity. 

§  265.  As  the  second  power  is  found  by  multiplying  the 
exponent  of  a  root  by  2,  (§  254,)  so  the  second  root  will  be 
found  by  dividing  the  exponent  of  a  power  by  2.  The  third 
root  will  be  found  by  dividing  the  exponent  of  the  power  by 
3.  And,  in  general,  any  root  may  be  found  by  dividing  the 
exponent  of  the  given  power  by  the  number  expressing  the 
root  to  be  found.  Thus,  the  second  root  of  a6  is  a3,  because 
cJxa8  =  a8.  The  third  root  of  a8  is  a3,  because  aaxa3X«a 
=  a6. 

Questions.  What  is  Evolution?  What  is  a  root  of  a  quantity! 
How  are  numeral  names  given  to  roots?  How  do  we  find  the  se- 
cond root  of  simple  quantities? 


§  268.3  EVOLUTION.  173 

§  266.  If  there  are  several  factors  in  the  quantity,  the  root 
of  each  factor  must  be  taken.  The  third  root  of  a9b3cB  is 
a3bc2 ;  for  a36c3  x  a3bc2  x  aBbca  =  a9b3c6. 

§  267.  If  the  quantity  have  a  numeral  co-efficient,  the  root 
of  the  co-efficient  must  be  taken  arithmetically.  The  second 
root  of  16asx6  is  4a*xa. 

§268.   The   root  of  a  fraction  is  found  by  taking  the 
root  of  both  numerator  and  denominator.     The  third  root  of 
8a6    .    2a3  2a9     2a*_     2a3  _  Sa^ 

27aT»  1S  3s5 ;  3x3  X3^  X  3^  ~~  27a;8 


EXAMPLES. 

1.  What  is  the  second  root  of  9a*? 

2.  What  is  the  third  root  of  SxB  ? 

3.  What  is  the  third  root  of  64b°  1 


■ii* 


4.  What  is  the  second  root  of  — ^-r  ? 

«4c6 

5.  What  is  the  second  root  of  8lx*y*1 

6.  What  is  the  third  root  of  27a366  ? 

7.  What  is  the  third  root  of  -^-  ? 

27x9 

8.  What  is  the  fourth  root  of  256a4x8  ? 

8  a3 

9.  What  is  the  third  root  of  — -*  ? 

125xe 

10.  What  is  the  fourth  root  of  S\ysxA2lsl 

1 6a*b* 

11.  What  is  the  second  root  of  --=- -_  0? 

49x6y2 

12.  What  is  the  second  root  of  a2b~*l 

13.  What  is  the  third  root  of  8a6c-flrf-3? 

14.  What  is  the  third  root  of  64ar3? 


Questions.     Supposing  the  quantity  has  several  factors  ?     A  nu- 
meral co-efficient  1     How  for  fractions  ) 
16* 


Ans. 

3as. 

Ans 

.  2x. 

Ans 

.  4b3. 

Ans. 

xy» 
a2c3 

Ans.  9x*y*. 

Ans. 

3ab\ 

Ans. 

2a3 
3x*' 

Ans. 

4ax». 

Ans. 

2a 
5x» 

Ans.  'Sy2xz3. 

Ans. 

4a2b 
lv»y 

Ans. 

ab~*. 

Ans.  2a2c 

-3d"\ 

ns.  4x  x  - 

4 

~~  x' 

174  ALGEBRA.  [e<*.  SEC.  15. 

§  269.  We  may  express  the  division  of  exponents  in  the 
same  manner  that  is  used  for  the  division  of  other  algebraical 

quantities.     Thus,  the  second  root  of  a*  is  a2  -  =  a2 ;  the 

fi  4 

third  root  of  a6  is  a7  -  =  a  ;  the  fourth  root  of  a*  is  a*  -  =  a1 

or  a.     By  the  same  rule,  the  second  root  of  a1  is  a?  ;  and  the 

l  1 

third  root  of  a1  is  «3  ;  the  second  root  of  a3  is  a2  ;  the  third 

2  l 

root  of  a9  is  a1 ;  the  nth  root  of  a  is  a*  ;  &c. 

"  ■       1 
§  270.  It  is  on  this  principle  that  the  exponent  *  is  now- 
used  as  the  sign  of  the  second  root,  and  the  exponent  7  is  the 
sign  of  the  third  root.*  * 

§  271.  Formerly,  the  sign  of  a  root  was  indicated  by  the 
radical  sign  </ ;  and,  for  some  purposes,  this  sign  is  still  in 
use.t     Whenever  it  is  used,  it  is  placed  to  the  left  of  the 
quantity ;  thus,  v/a.     The  number  of  the  root  is  denoted  by 
a  little  figure  placed  over  the  radical  sign;  unless  it  is  the 
second  root,  when  the  figure  2  is  omitted.     Thus, 
y/a  is  the  second  or  square  root  of  a. 
$fa  is  the  third  or  cube  root  of  a. 
Z/a  is  the  nth  root  of  a. 
\/a  +  x  is  the  second  root  of  (a -fa*.) 
IC7*  It  must  be  remembered,  that  if  the   radical  sign  is 
to  effect  more  than  one  factor,  the  vinculum  must  be  used 
with  it ;  thus,  \/5a,  V  ab.     When  the  radical  sign  is  placed 
before  a  fraction,  it  must  embrace  both  terms,  because  the 

line  in  the  fraction  forms  a  vinculum ;    thus,  */—  or      — . 

a        \a 

y/—  is  not  the  same  as  — — .     In  the  first  case,  it  is  the 


Questions.     How  may  exponents  be  divided?     What  signs  have 
been  derived  from  this  fact?     What  other  sign  for  roots?     Exan 
pies  ?     How  is  it  used  with  fractions  ? 

*  This  method  was  introduced  by  Simon  Stevinus,  of  Holland, 
about  1585. 

|  Invented  by  Stifelius,  in  1544. 


5  275.]  EVOLUTION.  175 

square  root  of  the  whole  fraction ;  in  the  last,  it  is  the  root 
of  the  numerator,  divided  by  7. 

§  272.  With  regard  to  the  signs,  a  positive  even  power  is 
formed  from  either  a  positive  or  negative  root ;  and  theretore, 
an  even  root  of  a  positive  quantity  is  either  positive  or  nega- 
tive. Thus,  the  second  root  of  a2  is  either  +«  or  — a ;  be- 
cause -f  a  x  -}-«  =  a2,  and  — a  x  — a  =  a*.  Hence  the  root 
is  said  to  be  ambiguous,  and  is  marked  with  both  signs,  ±a. 
The  second  root  of  2x  is  ±  \^2x. 

§  273.  The  odd  root  of  any  quantity  will  have  the  same 
sign  as  that  quantity.     The  third  root  of  — a3  is  — a. 

§  274.  There  cannot  be  an  even  root  of  a  negative  quan- 
tity;  because  no  quantity  can  be  multiplied  into  itself  an  even 
number  of  times,  in  such  a  manner  as  to  make  a  negative 
power.     §255.     See  also  §289. 

§275.  There  are  many  quantities  whose  roots  cannot  be 
exactly  found.  Thus,  the  second  root  of  2  can  never  be 
found.  The  root  is  then  expressed  either  by  the  radical  sign, 
or  a  fractional  exponent;  and  called  a  surd,  or  irrational 
quantity.  The  second  root  of  6  is  ^/6;  the  second  root  of 
x  is  y/x  or  x* ;  the  third  root  of  a -f-  x  is  (a+#)7  or  x^a-fa?. 

EXAMPLES. 

Express  the  roots  of  the  following  quantities,  first  by  the 
radical  sign,  and  then  by  the  fractional  exponents. 

15.  The  second  root  of  ax  1  Ans.  V ax  or  (ax)2. 

16.  The  third  root  of  a—yl       Ans.   *&a—y,  or  (a—  y)3. 

17.  The  second  root  of  a2x  ? 

/ l  i. 

Ans.   v  a2x,  or  a^/x,  or  (a2x)2,  or  ax^. 

18.  The  third  root  of  (a2— a) ;  and  the  third  root  of  x*1 

19.  The  second  root  of  a3 ;  and  the  second  root  of  ay3 1 

Questions.  Are  roots  positive  or  negative  1  Supposing  the  name 
of  the  root  is  an  odd  number  %    What  are  surds  1 


176  ALGEBRA.  [EQ.  SEC.  15. 


THE   SQUARE    ROOT    OF    BINOMIALS. 

§  276.  We  have  shown,  §  179,  that  the  second  power  or 
square  of  a+b  is  a'i-\-2ab-\-b2;  and  that  the  second  power 
of  a—b  is  a3 — 2ab+b*.  For  distinction's  sake  we  call  a-f  b 
a  binomial,  and  a— »6  a  residual.* 

§277.  Now,  the  second  power  of  a  binomial  consists  of 
three  terms,  the  first  and  last  of  which  are  complete  powers, 
and  the  middle  one  is  twice  the  product  of  the  roots  of  the 
two  powers;  and  all  connected  together  by  the  sign  -f-« 
Therefore,  whenever  we  meet  with  any  such  quantity,  we 
know  immediately  the  second  root  of  it  is  a  binomial,  and 
is  found  by  taking  the  roots  of  the  two  terms  which  are 
complete  powers,  and  connecting  them  by  the  sign  -f . 

§  278.  The  second  power  of  a  residual  is  altogether  like 
that  of  a  binomial,  excepting  the  sign  before  the  middle  term 
is  — .  The  root  of  such  a  quantity  is  taken  just  as  if  it 
were  a  binomial,  only  making  the  connecting  sign  a  —  mi- 
nus instead  of  a  plus. 

EXAMPLES. 

1.  What  is  the  second  root  of  a9+2ax-\-x2i.  Ans.  a  +  .r. 

2.  What  is  the  second  root  of  x*+2x+ 1?  Ans.  x  +  l. 

3.  What  is  the  second  root  of  4x* — 4ax+a*l  Ans.  2x — a. 

4.  What  is  the  second  root  of  x*  —  x+±  ?  Ans.  x — £. 

5.  What  is  the  second  root  of  x*-\- xy  +  —  ?      Ans.  x-\-~. 

*      4  2 

6.  What  is  the  second  root  of  a  +  2y/a~b  +  bl  Ans.  a?  +  b*. 

7.  What  is  the  second  root  of  16a3—  \6ab-\-4b'2l 

Ans.  4a — 2b. 

Questions.  What  is  the  difference  between  a  binomial  and  residual  ? 
How  do  we  know  when  the  second  root  of  a  quantity  is  a  binomial  ] 
How  is  it  found  ?  How  is  a  residual  root  found  ?  How  can  you 
tell  whether  a  trinomial  quantity  is  a  perfect  square'? 

*  These  terms  were  introduced  by  Dr.  Recorde,  in  1557. 


§281.]    EXTRACTION  OF  THE  SECOND  ROOT  OF  NUMBERS.       171 


EXTRACTION  OF  THE  SECOND  ROOT  OF  NUMBERS. 

§  279.  From  the  principle  last  shown,  we  derive  a  rule  foi 
finding  the  second  root  of  numbers.  For,  supposing  we  have 
a  number  of  two  digits,  say  54 ;  it  may  be  represented  by 
a-{-b  ;  that  is,  50  may  be  expressed  by  a,  and  4  by  b,  so  that 
50  and  4=a-\-b.  We  have  then  the  equation  a-f-^  =  54; 
a  equaling  the  tens,  and  b  equaling  the  units. 
Involving  both  members,  we  have  a2 -\-2ab +b2  ■=  2916. 

§280.  Now,  if  any  of  my  pupils  should  see  this  last 
equation  without  being  told  what  was  the  root  we  first  em- 
ployed, he  would  immediately  know,  by  §277,  that  the 
second  root  of  the  first  member  is  a +b,  and  of  course,  the 
root  of  2916  will  also  consist  of  two  quantities  that  may  be 
represented  by  a  and  b.  Thus,  v/2916  =  a-f&. 

It  would  be  only  necessary  to  inform  him  that  a  stands  for  an 
even  number  of  tens,  and  b  stands  for  a  quantity  less  than  ten. 

§  281.  As  it  is  known  that  the  second  power  of  10  is  100, 
the  second  power  of  20  is  400,  the  second  power  of  80  is 
6400,  &c. ;  it  is  evident  that  a  (which  stands  for  tens)  must 
be  the  root  of  the  hundreds.  Therefore,  to  find  a,  is  simply 
to  find  what  is  the  greatest  second  power  of  29  in  the  hun- 
dred,  and  then  to  take  the  root  of  it.  The  second  power  is 
25  hundred,  and  the  root  is  5;  for  6  times  6  =  36,  which  is 
too  much.  But  the  root  of  hundreds  is  in  tens ;  therefore, 
a  =  50,  and  a2  =  2500. 

Questions.  From  what  principle  do  we  derive  a  rule  for  extracting 
the  second  root  of  numbers'?  In  what  manner  may  a  number  be 
similar  to  a  binomial?  Which  letter  represents  the  tens?  How 
many  figures  in  the  power  are  represented  by  b  1  How  can  we  find 
the  second  root  of  a  number  represented  by  a21  And  then  the  value 
of  a?  Then  what  part  of  the  binomial  square  is  left  1 
M 


178  ALGEBRA.  £EQ.  SEC.   15. 

§  282.  Taking  a8  from  the  first  member  of  the  above  equa- 
tion, and  2500  from  the  second,  we  have  left  the  equation 

2a6+68  =  2916— 2500 -=416. 
As  a  =  50,  2a  =  100  ;  hence,  substituting  the  value  of  2a, 
we  have  1006+63  =  416.     And,  of  course,  if  we  divide  416 
by  100,  we  shall  find  very  nearly  the  value  of  6.     416-7-100 
=  4,  which  we  will  suppose  to  be  the  value  of  6. 

§  283.  Let  us  see  whether  b  does  actually  equal  4.  Our 
last  equation  was  1 006 + 68  =  4 1 6 

But  1006+68  =  (100  +  6)  Xb.     See  §  196. 
Substituting  (100+6)  x6,  for  1006  +  68,  (100+6)x6  =  416. 
If  6  =  4,  then  (100+4)  x4  should  =416. 

which  is  the  fact.     And  therefore,  6  does  equal  4.     Hence, 
a+6  =  50  +  4  -  =  54,  the  second  root  of  2916. 

§  284.  We  have  now  algebraically  discovered  the  following 
rule  for  extracting  the  second  root  of  any  number  that  con- 
sists of  not  more  than  four  figures.  Find  the  greatest  second 
power  of  hundreds,  and  put  its  root  in  the  answer.  Sub- 
tract that  second  power  from  the  whole  number,  and  divide 
the  remainder  by  twice  the  root  already  found.  Suppose  the 
quotient  to  be  the  number  that  completes  the  root;  add  it  to 
the  root  already  found,  and  also  to  the  last  divisor.  Mul- 
tiply the  divisor  thus  augmented,  by  the  quotient  figure  last 
found,  and  subtract  the  product  from  that  remainder  which 
was  divided.  If  there  is  nothing  left,  you  have  found  the 
true  root. 

§  285.  It  sometimes  happens  that  the  product  which  is 
obtained  by  multiplying  the  augmented  divisor,  is  greater  than 
the  number  which  was  divided.  In  such  cases,  the  figure  last 
put  in  the  root  is  too  large ;  and  therefore  it  must  be  erased, 
and  a  smaller  figure  substituted. 

Questions.  Of  which  factor  can  we  find  the  value?  How? 
What  can  we  do  with  this  value  ?  What  is  the  rule  for  extracting 
the  root  as  now  explained?  What  little  alteration  is  sometimes 
required  ? 


§287.]    EXTRACTION  OF  THE  SECOND  ROOT  OF  NUMBERS.        179 


7396(80 
6400 

160)996(6 
166)996 

Ans. 

68 

Ans. 

72 

Ans. 

23 

Ans. 

42 

Ans. 

98 

Ans. 

48 

EXAMPLES. 

1.  What  is  the  second  root  of  7396  ? 
Ans.  a2  must  equal  6400 ;  and  a1  =  80. 

Therefore  the  remainder  2ab  +  b*  =  996. 
2a  =  160,  which  is  contained  in  996,  6 
times ;  for  b  times  (2a  +  b),  or  6  times 
160  +  6  =  996.  Therefore  the  second  root 
of  7396  =  86. 

2.  What  is  the  second  root  of  4624  ? 

3.  What  is  the  second  root  of  5184? 

4.  What  is  the  second  root  of  529  ? 

5.  What  is  the  second  root  of  1764  ? 

6.  What  is  the  second  root  of  9604  ? 

7.  What  is  the  second  root  of  2304  ? 

§286.  It  will  be  found  by  trial,  that  if  a  number  is  in 
units,  its  second  power  will  be  in  tens;  for  the  square  of 
9  =  81.  If  the  number  is  in  tens,  its  second  power  will  con- 
tain more  than  two  figures,  and  less  than  five;  for  it  will  take 
the  square  of  100  to  make  10000.  The  principle  is  general, 
that  every  figure  in  the  root  except  the  left-hand  one,  re- 
quires two  figures  in  the  second  power;  and  for  the  left- 
hand  figure,  there  may  be  either  one  or  two  in  the  power. 

Hence,  by  pointing  off  the  number  from  the  right,  in 
periods  of  two,  (as  in  the  next  example,)  we  may  learn  how 
many  figures  will  be  in  the  root. 

§  287.  When  it  is  found  that  the  root  will  consist  of  more 
than  two  figures,  we  are  to  suppose  at  first  that  a3  is  repre- 
sented by  the  first  period ;  and  after  the  first  figure  in  the 
root  is  found,  the  first  two  periods  represent  a2 ;  and  so  period 

Questions.  How  many  figures  in  the  power  belong  to  each  figure 
in  the  root  ?  How  then  shall  we  learn  the  number  of  figures  in 
the  root?  How  is  the  operation  performed  when  the  root  contains 
more  than  two  figures  1 


180 


ALGEBRA. 


[EQ.  SEC.  15 


tfter  period,  a  represents  that  part  of  the  root  which  is  found, 
and  b  embraces  all  which  remains  to  be  found. 


8.  What  is  the  second  root  of  55225  ? 


55225(200 
40000 


400 
_£0 

4J0 

460 


5225(30 


Cl5i 
1 1290( 


Operation.     At  first,  a  =  200,  and 
b  =  30.      Afterwards,   a  =  230,   and 

1  C  2325(5 
m  I  2325 

§  288.  The  operation  may  be  abbreviated  by  omitting  the 
ciphers  in  the  roots  and  powers,  as  in 
the  margin.  When  we  do  this,  how^ 
ever,  we  must  remember  that  a  repre- 
sents an  order  ten  times  as  great  as 
that  represented  by  b ;  and  therefore, 
when  we  double  it  for  a  divisor,  we 
must  annex  a  cipher  to  it,  to  bring  it 
to  the  same  denomination  as  b. 

9.  What  is  the  second  root  of  2125764  ? 

10.  What  is  the  second  root  of  20736  ? 


55225(235 
4 


^$152 
«!  J  129 

«°  C  2325 
m  I  2325 


11.  What  is  the  second  root  of  10342656? 

12.  What  is  the  second  root  of  36372961  ? 


Ans.  1458. 
Ans.  144. 
Ans.  3216. 
Ans.  6031. 


§  289.  If  the  number  to  be  evolved  is  a  vulgar  fraction,  it 
must  be  reduced  to  decimals,  which  are  to  be  pointed  off  from 
units  towards  the  right,  instead  of  towards  the  left. 

Thus,  v/J?  —  ^.25  -  =  .5     v/f  =  ^.428571428,  &c. 

It  will  be  pointed  off  thus,  .42857142,  &c.     The  pupil  will 
find  the  answer  to  be  .65465  +  . 


Question.     Suppose  the  power  is  a  vulgar  fraction  1 


§  293.]         PURE  QUADRATIC  EQUATIONS.  1R1 

SECTION  XV. 

PURE  QUADRATIC  EQUATIONS. 

§  290.  There  are  several  kinds  of  equations.  We  have 
hitherto  attended  to  that  kind  which  contains  an  unknown 
quantity  in  the  first  degree  only.  But  there  are  equations 
that  contain  an  unknown  quantity  in  the  second  power ;  and 
also  in  the  third  power,  and  so  on  indefinitely. 

§291.  In  order  to  distinguish  the  different  kinds  of  equa- 
tions, we  are  accustomed  to  call  those  which  contain  only  the 
first  power  of  the  unknown  quantity,  simple  equations,  or 
equations  of  the  first  degree.  Those  which  contain  the  se- 
cond  power  of  the  unknown  quantity,  are  called  quadratic 
equations,  or  equations  of  the  second  degree.  Those  which 
contain  the  third  power  of  the  unknown  quantity,  are  called 
equations  of  the  third  degree,  &c. 

§  292.  There  are  two  kinds  of  quadratic  equations.  Those 
which  contain  only  the  second  power  of  the  unknown  quan- 
tity, are  called  pure  quadratic  equations ;  those  which  contain 
both  the  second  and  the  first  power  of  the  unknown  quantity, 
are  called  affected  quadratic  equations.* 

§  293.  A  "pure  quadratic  equation  is  reduced  by  simply 
extracting  the  root  of  both  members. 

Questions.  How  do  we  distinguish  different  kinds  of  equations'? 
How  many  kinds  of  quadratics'?  How  do  we  reduce  a  pure  quad- 
ratic equation  ? 

*  All  pure  equations  are  called  by  some  algebraists  simple 
equations. 

16 


182  ALGEBRA.  [SECTION    XV. 


EQUATIONS.— SECTION  15. 

1.  There  is  a  number,  such,  that  by  adding  5  to  it  for  one 
factor,  and  subtracting  5  from  it  for  another  factor,  we  may 
obtain  96  for  the  product.     What  is  that  number  ? 

Stating  the  question,  Let   x  =  the  number. 

a?+5  =  one  factor. 

x — 5  =  the  other. 
(x+5)x(#— 5)  =  the  product 
Forming  the  equation,  (a?+ 5)  x  (a?—  5)  =  96 

Multiplying,  §  172,  a:3— 25  =  96 

Transposing  and  uniting,  xa  =  121 

Extracting  the  root,  x  =  ±  1 1. 

ICT"  In  this  example,  we  have  given  1 1  the  uncertain  sign, 
because  121  is  the  second  power  of  either  +11  or  — 11.  But 
the  x*  we  know  to  be  the  second  power  of  +#,  because  x 
was  +  in  the  supposition.  We  have  therefore  found  that 
the  required  number  is  either  11,  or  11  less  than  nothing. 
Both  values  will  satisfy  the  conditions  of  the  question.  If 
11  =  #,  then  ll-f5=16,  and  11  —  5  =  6.  And  16x6  =  96. 
If  — ll=;r,  then  —11  +  5  =  6,  and— 11— 5= -16.  And 
6x  — 16  =  — 96. 

§  294.  In  this  manner,  every  quadratic  equation  will  have 
two  answers;  but  the  conditions  of  the  question  will  generally 
determine  which  answer  is  the  one  required.  In  pure  quad- 
ratics, we  generally  suppose  the  positive  answer  to  be  the 
true  one. 

2.  What  two  numbers  are  those  which  are  to  one  another 
is  3  to  5 ;  and  whose  squares,  added  together,  make  1666? 

Questions.  Why  db  in  the  answer  ?  How  many  answers  then 
iii  quadratics  ? 


$  294.]  PURE    QUADRATIC    EQUATIONS.  183 

Stating  the  question, 


Forming  the  equation, 

Multiplying  and  uniting, 
Dividing  by  34, 
Extracting  the  root, 


x  = 

the  greatest. 

Sx 

1F  = 

the  least. 

x»  = 

square  of  the  greatest. 

9x* 
25"  = 

:  square  of  the  least. 

.     9#a 

x2  4 = 

^25 

:1666 

34z3  = 

:41650 

#3  = 

=  1225 

x  = 

:±35  the  greatest. 

|  of  35  =  21  the  least. 


3.  The  distance  to  a  certain  place  is  such,  that  if  96  be  sub* 
traded  from  the  square  of  the  number  of  miles,  the  remainder 
will  be  48.     What  is  the  distance  ?  Ans.  12  miles. 

4.  There  is  a  field  containing  108  square  rods;  and  the 
sum  of  the  length  and  breadth  is  equal  to  twice  their  differ- 
ence.    Required  the  length  and  breadth  ? 

108 
ICT*  In  stating  the  question,  x=  the  length,  and  = 

the  breadth;  because  the  two  multiplied  together  make  108, 
the  area.  In  this  operation,  there  will  be  an  equation  having 
— xz  for  one  member.  But  as  by  §  274,  we  cannot  find  the 
second  root  of  a  negative  quantity,  we  must  first  change  the 
signs  of  all  the  terms.  Ans.  Length,  18  ;  breadth,  6. 

5.  There  is  a  rectangular  field,  whose  breadth  is  £  of  the 
length.  After  laying  out  £  of  the  whole  ground  for  a  garden, 
it  was  found  that  there  were  left  625  square  rods  for  mowing 
Required  the  length  and  breadth  of  the  field  ? 

Ans.  Length,  30  rods  ;  breadth,  25. 

6.  Two  men  talking  of  their  ages,  one  said  that  he  was  94 
years  old.  Then,  replied  the  younger,  the  sum  of  your  age 
and  mine,  multiplied  by  the  difference  between  our  ages,  will 
produce  8512.     What  is  the  age  of  the  younger? 

Ans.  18  years* 


184  ALGEBRA.  [SECTION  XV. 

7.  If  three  times  the  square  of  a  certain  number  be  divided 
by  4,  and  if  the  quotient  be  diminished  by  12,  the  remainder 
will  be  180.     What  is  the  number  ?  Ans.  16. 

8.  A  person  bought  a  quantity  of  sugar  at  such  a  rate,  that 
the  price  of  a  pound  was  to  the  number  of  pounds  as  4  to  5. 
If  the  cost  of  the  whole  had  been  45  cents  more,  the  numbei 
of  pounds  would  have  been  to  the  price  of  a  pound  as  4  to  5. 
How  many  pounds  did  he  buy,  and  what  was  the  price  per 
pound  ? 

Operation. — Let  a?  =  number  of  pounds,  and  y  =  the  price, 

and  xy  =  the  whole   cost.     Then,  by  the   first  condition, 

4x 
y  =  — .    If  x  pounds  cost  xy-\- 45,  then  one  pound  would  cost 
o 

-^ .   Therefore,  by  the  second  condition,  x  =  — — ; 

x  5a? 

4a?  1 6a?3 

or.  5a:9  =  4zi/ +180.     But  y  =  —  ;.•.  4xy=-—.    Whence 

16a?9 
the  equation  5a?9  mm  — - — (-180. 

Ans.  10  pounds,  at  8  cts.  a  pound. 

§  295-  The  pupil  will  find  that  whenever  both  unknown 
quantities  are  found  in  one  term,  the  second  method  of  ex- 
termination (§131)  is  the  best. 

9.  There  are  two  numbers  whose  product  is  144,  and  the 
quotient  of  the  greater  by  the  less  is  16.  What  are  the  num- 
bers ?  Ans.  48  and  3. 

10.  What  two  numbers  are  those,  whose  sum  is  to  the 
greater  as  11  to  7;  the  difference  of  their  squares  being  132? 

Ans.  14  and  8. 


§  300.]       AFFECTED  QUADRATIC  EQUATIONS.  185 

SECTION   XVI. 
AFFECTED  QUADRATIC  EQUATIONS.  ' 

§  296.  It  is  but  very  rarely  that  we  find  quadratic  equa- 
tions so  simple  as  those  in  the  preceding  section.  Most  gene- 
rally, when  reduced  to  their  simplest  terms,  they  contain  the 
second  power  of  the  unknown  quantity  in  one  term,  the  first 
power  of  the  unknown  quantity  in  another  term,  and  a  known 
quantity  as  a  third  term;  thus,  a?9+2a?==24. 

§  297-  Such  equations  are  called  affected  quadratic  equa- 
tions, because  the  operation  on  the  second  power  is  affected 
by  a  different  power  of  the  same  quantity.* 

§  298.  It  is  of  no  importance  how  many  terms  are  found 
in  an  equation,  provided  they  are  the  first  and  second  powers 
of  the  unknown  quantity,  and  known  numbers ;  because  they 
can  all  be  united  so  as  to  form  but  three.  Thus,  x-\-&—2x* 
-f3a?-f-10  =  5a?3,  can  be  united  so  as  to  form  — 7a?2-f-4a?  = 
—  18;  or  7a?3— 4x  =  18. 

§  299.  Equations  of  this  kind  would  seem  at  first  to  be 
very  difficult  of  solution ;  but  by  the  application  of  a  principle 
which  is  found  in  raising  a  binomial  to  its  second  power,  and 
evolving  it  again,  we  manage  them  with  very  little  trouble. 

§  300.  Our  object,  then,  in  the  first  place,  is  to  form  from 
these  two  terms  which  contain  the  unknown  quantity,  a  com- 
plete second  power  of  some  binomial.  Of  what  quantities 
this  binomial  is  composed,  we  care  not;  say  a?3-j-2aa?+«3. 

Questions.  What  is  the  most  general  form  of  quadratics  ?  "What 
are  such  equations  called  ?  How  many  terms  may  they  contain  % 
What  is  the  first  procedure  in  solving  such  equations  ? 

*  This  name  was  introduced  by  Vieta,  about  the  year  1600.  It 
is  derived  from  the  Latin  affedo,  to  pester  or  trouble. 

16* 


186  ALGEBRA.  [SECTION  XVI. 

§  301.  Now,  the  second  power  of  a  binomial,  (see  §  179,) 
has  in  its  jirst  term,  the  second  power  of  the  first  term  in  the 
binomial.     Thus, 

Binomial,  x-\-a. 

Second  power,  #s+2ax-|-a3. 
For  this,  we  use  the  second  power  of  the  unknown  quantity 
in  the  equation,  say  x*. 

§  302.  Again,  the  second  power  of  the  binomial  has  in  its 
second  term,  the  first  power  of  the  first  term  in  the  binomial 
as  one  factor ;  and  twice  the  last  term  of  the  binomial  as  the 
other  factor.     That  is,  it  has  2a  times  x.     Thus, 
Binomial,  x+a 

Second  power,  x9-f2aa?-|-a9. 
For  this,  in  our  equation,  we  use  the  term  that  has  x  the  first 
power,  in  it.     And  we  also  determine,  that  whatever  its  co- 
efficient is,  that  co-efficient  shall  represent  the  value  of  2a. 

§  303.  Finally,  the  last  term  in  the  second  power  of  the 
binomial  is  the  second  power  of  half  that  other  factor  which 
accompanies  x  in  the  second  term;  that  is,  it  is  the  second 
power  of  once  a.  Therefore,  in  our  equation,  whatever  was 
the  co-efficient  in  the  second  term,  the  second  power  of  half 
that  co-efficient  will  be  the  last  term  which  it  is  necessary  to 
add,  in  order  to  make  up  the  complete  second  power  of  a 
binomial. 

EXAMPLES. 

1.  Suppose  we  have  the  equation  x*+4x  =  96. 

ICT"  We  are  first  to  make  the  left-hand  member  similar  to 
the  formula  x3-{-2ax+ a3.  The  co-efficient  of  the  second  term 
in  the  question  is  4 ;  and  therefore  4  represents  the  2a  in  the 
formula.     The  half  of  4,  which  is  2,  will  represent  a :  and 

Questions.  In  this  operation,  what  part  of  the  binomial  square  do 
we  make  of  x2  ?  What  do  we  consider  as  represented  by  the  co- 
efficient of  x  in  the  second  term  ]  Why  1  How  do  we  obtain  the 
last  term  of  the  binomial  square?     Why  is  this  correct? 


§  303.]      AFFECTED  QUADRATIC  EQUATIONS.  187 

the  second  power  of  2,  which  is  4,  will  represent  a2.  There- 
fore, from  the  left-hand  member  of  the  equation  we  may  form 
the  complete  second  power  x2-{-4x-\- 4.  But  if  we  add  4  to 
this  member,  we  must  also  add  4  to  the  right-hand  member, 
to  preserve  the  equation.     We  shall  then  have  the  equation 

a?3+4#+4  =  96-f4 
Or,  #3-+4x-f4  =  100 

Extracting  both  sides,  §277,  284,      #-f  2  =  ±10 
Transposing  2,  x  =  8  or  — 12. 

Proof.     If  x  =  8,  then  x2+4x  =  64  +  32  -  =  96. 
If  x=— 12,  then  x2+4x  =  144— 48-  =  96.    For,  it  must  be 
remembered  that  if  x  is  — 12,  then  4x  =  — 48,  which  is  to 
be  added  to  144;  and  adding  — 48,  is  done  by  writing  it  im- 
mediately after  the  144  with  its  own  sign. 

2.  Given  x2-\-6x=  16,  to  find  the  value  of  x. 

Here,  6  =  2a  in  the  formula.  .*.  a  =  3,  and  a2  =  9.   There- 
fore, completing  the  square,  #9-f-6.r+9  =  16-f  9-  =  25. 
Extracting  the  root,  x-\- 3  =  ±5 
Transposing,  x  =  2  or  — 8. 

3.  Given  rr3+8x  =  33. 

Half  the  co-efficient  of  the  second  term  is  4 ;  and  the  second 
power  of  4  is  16.     Therefore, 
the  square  is  completed  thus,  x2-{-Sx-\-\&  =  33-f  16. 

Ans.  x  =  'S  or  — 11. 

4.  Given  #3-f  4#  =  32,  to  complete  the  square. 

Ans.  Half  the  co-efficient  of  the  second  term  is  2  ;  and  the 
second  power  of  2  is  4.  Therefore,  the  square,  when  com- 
pleted, is  x2 +4x  +  4  =  32  +  4.     And  x  —  4  or  —8. 

5.  Given  x2 — 4x  =  32,  to  complete  the  square.  . 
Operation. — The  left-hand  member  of  this  equation  is  a 

part  of  the  second  power  of  a  residual  quantity.  See  §  276. 
Now,  as  the  second  power  of  a  residual  is  the  same  as  that 
of  a  binomial,  with  the  exception  of  the  second  term  being  — 


188  ALGEBRA.  [SECTION  XVI. 

instead  of  + ,  the  square  is  completed  in  the  same  manner  as 
before.  Thus,  half  of  the  co-efficient  is  2,  and  the  second 
power  of  2  is  4.  Therefore,  when  the  square  is  completed, 
it  is  x*— 4x+4  =  32  +  4-=36 

Extracting  the  root,  see  § 278,  .   x— 2  =  ±6 

And  transposing,  a?  =  8  or — 4. 

6.  Given  Xs— Gx  =  —8,  to  find  x. 
Completing  the  square,  x* — 6:r+9  =  — 8-f-9  -  =1 

Extracting  the  root,  x — 3  =  ±  1 

Transposing,  x  =  2  or  4 

Complete  the  square  in  the  following  equations,  and  find 
the  value  of  x. 


7. 

a;a=4;r  =  140. 

Ans. 

07=  10  or  ■ 

-14, 

8. 

a-s+8x  =  65. 

Ans. 

a?  =  5  or  - 

-13. 

9. 

a:"— 4a?  =  45. 

Ans. 

x  =  9  or  — 

-5. 

10. 

x*—  15a?  =  —54. 

The  half  of  15  is  — 
2 

;  anc 

completing  the  square, 

X*- 

1R       225 
-15a?+—  =  - 

-54+—. 

Or, 

X*- 

i*     ,225 
-15a?+— =- 

216     225 

-4T+— 

9 

~4 

Extracting  the  root, 

15 

x =  ±- 

2       ^2 

Transposing,  x  =  9  or  6. 

11.  x*-\-x  =  42.  In  this  example,  half  of  the  co-efficient 
is  £.  Ans.  a?a+a?+£  =  42£.     And  x  =  6  or  —7. 

12.  2a?3+16a?  =  40. 

We  must  remember  that  the  first  term  in  the  second  power 
of  the  binomial  is  a?3.  On  this  account,  if  a?3  in  the  equation 
has  a  co-efficient,  we  must  divide  the  equation  by  that  co- 
efficient.    This  example  then  becomes 

a?3+8a?  =  20.  Ans.  x  =  2  or  —10. 

13.  5a?3— 60a?  =  800.  Ans.  x  =  20  or  —8 


§304.]       AFFECTED  QUADRATIC  EQUATIONS.  189 

14.  8a?3 -f  8a?  =16.    (See  Ex.  11.)     Ans.  a?  —  1  or  —2. 

5  1 

15.  6a?3 — 5a?  =  l.  When  divided,  it  is  a?2 — —  a?=— . 

6  6 

Half  of  -  =  — ,  which  squared  =  — — . 
6       12  144 

.    .  5        25        24       25  49 

Completing  the  square,    *»-  -  x+  —  =  —  +  —  -  =  — . 

5  7 

Extracting  the  root,  x =  ± — • 

6  12  12 

2 
Transposing,  x  =  1 ,  or  —  — . 

12 

t/>      a;3      a;      133  ™  i  •  v   ,   •   .       „     2a?       133 

16.  _  —  -  =  —.  Multiplied,  it  is  a?2—— =  —. 

n        .    .     ■  _      2        1       133      1         400 

Completing  the  square,  a?3—  -a?+-  =  -—+--  =  — . 

1       20 

Extracting  the  root,  x =  — . 

S         3 

Transposing,  x  =  7  or  — 6J. 

17.  3a?3— 3a:  =  —  f.     Divided,  it  is  a?3— a?  =  — f.     And 

18.  14a?— a?3  =  45.     Transposed,  it  is  —  a?3  +  14a?  =  45. 
The  first  term  of  the  second  power  of  a  binomial  is  positive. 

We  must,  therefore,  change  the  signs  of  the  whole  equation, 
§  62,  in  order  to  make  the  first  term  of  the  equation  positive. 
It  will  then  be  a?3 — 14a?  =  — 45.  Ans.  a?  =  9  or  5. 

In  review  of  the  foregoing  principles,  we  establish  the  fol- 
lowing rule. 

RULE    FOR   THE    SOLUTION    OF    QUADRATIC    EQUATIONS. 

§  304.  Unite  and  transpose  the  terms  of  the  equation  in 
such  a  manner  that  the  second  power  of  x  shall  occupy  the 
first  term  of  the  equation,  and  be  positive.  Let  the  first 
power  of  Jibe  in  the  second  term,  and  all  the  known  quan- 
tities be  in  the  right-hand  member. 


190  ALGEBRA.  [SECTION  XVI 

§  305.  Then,  if  the  first  term  has  any  co-efficient,  divide 
the  whole  equation  by  that  co-efficient. 

To  both  sides  of  the  equation,  add  the  second  power  of 
half  the  co-efficient  of  the  second  term;  and  the  unknown 
side  will  then  be  a  complete  square.  Extract  the  second 
root  of  each  side,  taking  care  to  prefix  the  double  sign  ± 
to  the  right-hand  member.  And  then,  by  transposition, 
the  value  of  the  unknown  quantity  is  easily  found. 

IC73"  In  this  kind  of  equations,  it  is  not  necessary  to  destroy 
the  fractions.   Indeed  we  are  sometimes  obliged  to  make  them. 

EQUATIONS.— SECTION  17. 

1.  What  two  numbers  are  those  whose  sum  is  12,  and 
whose  product  is  35  ? 

Stating  the  question,  x  =  the  greater  number. 

12 — x=  the  less. 
\2x — x*  =  the  product. 
Forming  the  equation,  \2x— x*  —  35, 

Transposing,  §  304,  a?3—  12a?  =  —35, 

Completing  the  square,    a?3—  12x-f  36  =  —  35-f  36-  =  1, 
Extracting  the  root,  x — 6  =  db  1 , 

Transposing,  x  =  7  or  5. 

2.  Find  two  numbers,  such,  that  their  difference  shall  be  *S, 
and  their  product  240. 

|C7»  The  numbers  are  x  and  x  +  8,  Ans.  12  and  20. 

3.  The  difference  of  two  numbers  is  4,  and  their  product 
96.     What  are  those  numbers  ?  Ans.  12  and  8. 

4.  There  are  two  numbers,  whose  difference  is  9;  and 
whose  sum,  multiplied  by  the  greater,  produces  266.  What 
are  those  numbers?  Ans.  5  and  14  ;  or  — 9£  and  — 18*. 

Questions.  What  is  the  first  step  in  the  solution  of  affected  quad- 
ratic equations  ?  What  if  the  first  term  has  a  co-efficient  T  How 
do  we  complete  the  square]  What  is  the  next  step  ?  What  are  we 
to  do  in  case  of  fractions  1 


§  305.]  AFFECTED    QUADRATIC    EQUATIONS.  191 

5.  Divide  the  number  56  into  two  such  parts,  that  their 
product  shall  be  640.  Ans.  40  and  16. 

6.  Divide  the  number  60  into  two  such  parts,  that,  their 
product  shall  be  864.  Ans.  36  and  24. 

7.  The  difference  of  two  numbers  is  6,  and  the  sum  of  their 
squares  is  50.     What  are  the  numbers  ?  Ans.  7  and  1. 

8.  The  ages  of  a  man  and  his  wife  amount  to  42  years, 
and  the  product  of  their  ages  is  432.  What  is  the  age  of 
each?  Ans.  Man's,  24  years  ;  wife's,  18  years. 

9.  A  merchant  has  a  piece  of  broadcloth  and  a  piece  of  silk. 
The  number  of  yards  in  both  is  110;  and  if  the  square  of  the 
number  of  yards  of  silk  be  subtracted  from  80  times  the  num- 
ber of  yards  of  broadcloth,  the  difference  will  be  400.  How 
many  yards  are  there  in  each  piece  ? 

Ans.  60  of  silk ;  50  of  broadcloth. 

10.  A  is  4  years  older  than  B;  and  the  sum  of  the  squares 
of  their  ages  is  976.     What  are  their  ages  ? 

Ans.  A's  age,  24  years ;  B's,  20  years. 

11.  A  nursery-man  planted  8400  trees  at  equal  distances, 
in  the  form  of  a  rectangle,  having  50  trees  more  in  front  than 
in  depth.     What  was  the  number  in  front? 

IC7"  Let  x  =  the  length,  and  x— 50  =  the  breadth. 

Ans.   120  trees. 

12.  Divide  the  number  30  into  two  such  parts,  that  their 
product  may  be  equal  to  eight  times  their  difference. 

Ans.  6  and  24. 

13.  A  person  being  asked  his  age,  answered,  My  mother 
was  20  years  old  when  I  was  born,  and  her  age  multiplied  by 
mine  exceeds  our  united  ages  by  2500.     What  was  his  age  ? 

Ans.  42  years. 

14.  The  length  of  a  certain  field  exceeds  its  width  by  8 
rods ;  and  its  area  is  768  rods.  What  are  the  dimensions  of 
the  field  ?  Ans.  Length,  32  rods ;  width,  24  rods. 


192  ALGEBRA.  [SECTION  XVI. 

15.  Divide  the  number  60  into  two  such  parts,  that  their 
product  shall  be  to  the  sum  of  their  squares  in  ihe  ratio  of 
2  to  5.  Ans.  20  and  40. 

16.  Each  of  two  captains  distributes  1200  dollars  equally 
among  his  men.  A  has  40  men  more  than  B,  and  B's  men 
receive  five  dollars  apiece  more  than  A's.  How  many  men 
had  each  captain  ? 

IQ0*  If  x  =  the  number  of  B's  men,  then =  the  share 

x 

of  each  of  them.     Also,  x+40  =  number  of  A's  men,  and 

— — — ■  =  the  share  of  each  of  them.     Ans.  A  had  120 ;  B,  80. 
x+40 

17.  A  man  bought  a  certain  number  of  sheep  for  80  dollars. 
If  he  had  bought  four  more  sheep  for  the  same  money,  they 
would  have  come  to  him  1  dollar  a  piece  cheaper.  How 
many  sheep  did  he  buy  ?  Ans.   16  sheep. 

18.  A  merchant  sold  a  quantity  of  cloth  for  $56,  by  which 
he  gained  as  much  per  cent,  as  the  whole  cost  him.  How 
much  did  it  cost  him  ? 

|C7*  If  x  =  cost,  then  56— x  =  gain.     Per  cent,  is  always 

x 

written  as  the  100th  of  a  number.     In  this  case,  it  is  » 

100 

and  the  cost  multiplied  by  it  makes  the  gain.  Ans.  $40. 

19.  A  person  buys  a  horse  for  a  certain  sum,  and  afterwards 
sells  him  for  $144,  and  gains  exactly  as  much  per  cent,  as  the 
horse  cost.     What  did  the  horse  cost  him  1  Ans.  $80. 

20.  A  person  bought  two  pieces  of  cloth  of  different  sorts  : 
of  which  the  finer  cost  4s.  a  yard  more  than  the  other.  For 
the  finer  he  paid  £18,  but  for  the  coarser,  which  exceeded 
the  finer  in  length  by  2  yards,  he  paid  only  £16.  How  many 
yards  were  there  in  each  piece  ? 

Ans.  18  yards  of  the  finer  :  20  of  the  coarser, 

THE    END. 


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